An examination of these tables reveals their purpose. They serve as an extremely quick and reliable method of calculating the positions of dates in the Long Count, and the Long Count values corresponding to given Calendar Round dates. The simplicity of the system lies in the fact that on adding units of the second order (364 x 20) the Calendar Round date is found by merely subtracting one month for each unit involved, or in the case of subtraction, one obtains the correct Calendar Round date by adding one month for each unit subtracted, without any change to the day sign or its coefficient.
Example 4:
9.10.10.0.0  13 Ahau 18 Kankin 
1.0.4.0  (364 x 20) Add 


9.11.10.4.0  13 Ahau 18 Mac (Mac is the month before Kankin) 
Example 5:
9.14.19.5.0  4 Ahau 18 Muan 
3.0.12.0  (364 x 20 x 3) Subtract 


9.11.18.11.0  4 Ahau 18 Cumhu (Cumhu is three months after Muan) 
In handling multiples of the first order (364 days) instead of subtracting
or adding whole months, one merely subtracts or adds one day to the month
column and advances the day sign four spaces without changing the coefficient
(Examples 2 and 3).
Example 6:
9.11.10.4.0  13 Ahau 18 Mac 
1.0.4  (364 x 1) Add 


9.11.11.4.4  13 Kan 17 Mac (Kan is 4 days forward from Ahau) 
Example 7:
9.11.18.11.0  4 Ahau 18 Cumhu 
12.2.8  (364 x 12) Add 


9.12.10.13.8  4 Lamat 6 Cumhu 
Lamat is 8 (4 x 12 = 48, 48  40 = 8) days forward from Ahau. 18 Cumhu drops 12 days to 6 Cumhu.
Fractions of 364 days are harder to handle. One keeps the day
sign coefficient and advances 11 days (91  20 x 4 = 11), as in the arrangement
of the table in Dresden page 32 (Example 1), and adds 91 days to the month
for each multiple of 91 days added.
Example 8:
9.11.11.4.4  13 Kan 17 Mac 
4.11  (91 days) Add 


9.11.11.8.15  13 Men 8 Cumhu 
Note that Men is next in the horizontal line to Kan on page 32, and
that 8 Cumhu is 4 months (80 days) and 11 days forward from 17 Mac.
Example 9:
9.12.10.13.8  4 Lamat 6 Cumhu 
13.13  (91 x 3 days) Subtract 


9.12.19.17.15  4 Men 13 Zotz 
Men is three places removed backwards in the horizontal line from Lamat on page 32 of the Dresden Codex. 13 Zotz is 13 months and 13 days back from 6 Cumhu.
Odd multiples of 91 days to 91 x 19 given in the table on pages 6364 of the Dresden Codex are short cuts in calculating positions in the 260day sacred almanac (the primary interest of the compilers and copyists of the Dresden Codex). Nevertheless, when using the Long Count in seeking Calendar Round dates it would have been simpler to have added, for example, five computing years and then subtracted 91 days rather than use 91 x 19 directly as given in this table. The arrangement of this and the other tables is ample justification for assuming vigesimal count in the computing year.
Having mastered this simple method, the Maya novice first arranges his grains of maize or pebbles, or possibly constructs a simple abacus with slots and pebbles. We will assume that he used this abacus, although the system without it would be exactly the same save that instead of rapidly flipping his pebbles, disks, or balls along a slot, he would pick them up one by one with his fingers. The abacus would be very similar in layout to that shown in figure 3. The lefthand side would have five divisions of equal size, each containing four horizontal slots. These divisions, placed one above the other, would represent cycles, katuns, tuns, uinals, and kins. Where the plan shows each partition labeled with the name in European characters, the Maya novice might have carved the corresponding glyph. Each slot would contain five counters, except the top slot of the uinal group, which would contain only three, since the coefficient of the uinal cannot exceed 18. Below the kin division three slots of somewhat larger size, holding five, five, and three counters in each mark the coefficients of the Long Count. At the top right, four slots with five counters in each mark the coefficients of the months, the names of which are immediately below in a vertical column with spaces to the left to hold the movable indicator. Across the bottom of the abacus the 20 day signs are arranged in the same sequence as in the Dresden Codex, so that one can move vertically or horizontally as in Examples 1 and 2. Space is left for the movable indicator to the left of each day sign, written out in our illustrations, but probably carved in wood in the original.
It is hard to say what the Maya novice would have used as indicators. We have no reason to suppose that he would share our use of a conventionalized arrow, and good grounds for saying that he would not have employed a hand with pointing finger. In the drawing of the abacus, a conventionalized sign for mat has been employed as day sign indicator. A mat symbolized authority or rule, and since each day was personified as a deity ruling over that period, the symbol is applicable, although there is no evidence that the Maya used it in that manner. The month sign indicator rests on firmer ground. It is based on an element found before month signs in Yucatecan inscriptions probably with a meaning such as “falls on” n day of month x. With his board thus completed and with al counters flipped to the left so that nothing registers on the right sides of all slots, the novice is ready for his calculations.
He has already memorized the minor vagaries of the computing year with such mnemonic aids as 411 (i.e. for one computing year add four days to one tun and move back one day in the month list; for one 20computingyear period add four uinals to one katun and move back one uinal in the month position). He then chooses and memorizes a date ending on the day 13 Ahau (the last day of the 260day sacred almanac) and uses that as the base for all his calculations throughout his life. The reason for choosing a date ending on 13 Ahau will appear in due course when reverse calculations are involved. A ver convenient date of this kind is 9.10.10.0.0, 13 Ahau 18 Kankin, since it is at the end of a half katun, and the uinal and kin coefficients are therefore zero. Accordingly, the apprentice flips over to the right side nine counters in the cycle division, 10 each in the katun and tun divisions, 13 in the day number division (for 13 Ahau) and 18 in the month number division (for 18 Kankin). The indicators are set at Ahau and Kankin respectively, and the abacus will appear as in figure 3a.
Example 10: The novice is told by the priest to calculate the day and month position of the Long Count date 9.16.12.5.17.
Step 1. A glance shows that 6 2ocomputingyear periods will bring
him fairly close to his target:
9.10.10.0.0  13 Ahau 18 Kankin 
6.1.6.10  (364 x 20 x 6) Add 


9.16.11.6.0  13 Ahau 18 Mol 
Accordingly he flips six counters across in the katun division, one in the tun division, six in the uinal division, moves the month indicator back six months (one month for every 364 x 20 period) and leaves the rest of the abacus untouched.
Step 2. His abacus now reads 9.16.11.6.0, 13 Ahau 18 Mol (fig.
3b). The addition of one computing year will bring him still closer
to the desired date:
9.16.11.6.0  13 Ahau 18 Mol 
1.0.4  Add 


9.16.12.6.4  13 Kan 17 Mol 
Accordingly, he flips across to the right one counter in the tun division and four in the kin division, moves the day sign indicator down one place in the day sign column, and takes away one counter (by flipping to the left) from the month number (as per Examples 6 and 7).
Step 3. His abacus now reads 9.16.12.6.4, 13 Kan 17 Mol (fig.
3c). He sees that he has overshot the mark by seven days:
9.16.12.6.4  13 Kan 17 Mol 
7  Subtract 


9.16.12.5.17  6 Caban 10 Mol 
He subtracts this by flipping to the left seven counters in the kin division (actually he must borrow one uinal in this case to subtract seven kins), seven counters in the day number, and seven counters in the month number division, and moves his day indicator back seven places in the straight 20 day count. The calculation is now complete, and his answer is that the day and month positions corresponding to 9.16.12.5.17 are 6 Caban 19 Mol (fig. 3d).
Example 11: Find the day and month positions of the Long
Count date 9.14.19.5.8. The moves from the same base are as follows:
9.10.10.0.0  13 Ahau 18 Kankin  
Step 1.  4.0.16.0  (364 x 20 x 4) Add 


9.14.10.16.0  13 Ahau 18 Yax  
Step 2.  8.1.12  (364 x 8) Add 


9.14.18.17.12  13 Eb 10 Yax  
Step 3.  4.11  (91 days) Add 


9.14.19.4.3  13 Akbal 1 Muan  
Step 4.  1.5  (25 days) Add 


9.14.19.5.8  12 Lamat 6 Pax 
Step 1. Flip across four counters in the katun division, 16 in the uinal division, and move the month indicator back four months.
Step 2. Flip across eight counters in the tun division, one in the uinal division, and 12 in the kin division. Move eight places in the vertical column from Ahau to Eb and flip eight counters backwards in the month number division. In adding the 11 counters to the kin number of 12, one proceeds as in “carrying” in our own arithmetic: 12 + 11 is 3 and carry 1. 17 uinals and 1 “carried” are 0, and “carry” 1 to the tun.
Step 3. The position sought is now less than one tun away but rather far to calculate in one’s head. Accordingly add one quarter of a computing year. Flip across four uinal and 11 kin counters. Move one place horizontally from Eb to Akbal, and add four months and 11 days to the month divisions (i.e. in this particular case move down five months and subtract nine days from the corresponding counters).
Step 4. Add the 25 days still lacking to reach the answer that 12 Lamat 6 Pax is the Calendar Round date corresponding to 9.14.19.5.8.
It will be noted that these calculations are as near mistakeproof as is possible. The day sign and coefficient, ritualistically the most important part of the Calendar Round date, remain unchanged through the larger part of the calculation. Other parts change in the simplest way possible—add one 20computingyear period; subtract one month. Add one computing year; subtract one day from the month, and in the first case leave the day sign and coefficient untouched; in the second, move the day sign one place.
Frequently one has the Calendar Round date, and it is necessary to find which position near a given katun of a given cycle it occupies in the Long Count. To achieve this one starts with the same base, 9.10.10.0.0, 13 Ahau 18 Kankin.
Let us suppose that the Long Count positioin of 8 Oc 13 Yax in the vicinity of 9.14.0.0.0 is needed.
Step 1. One first seeks the nearest Ahau day to 8 Oc. This is
5 Ahau 3 Zac ten days later. There then enters a simple formula which
one would think the Maya might have stumbled on, although it may savor
somewhat of European arithmetic. To reach any day Ahau from 13 Ahau
double the coefficient of the day Ahau to be reached (subtracting 13 if
the number is greater) and add the corresponding number of uinals.
As the day we are seeking is 5 Ahau, add 10 uinals, and then subtract 10
days to get back to the day Oc. If this formula were unknown to the
Maya, a table of the sacred almanac, like the fragmentary one on a Uaxactun
fresco, would give the same result with very little more trouble.
9.10.10.0.0  13 Ahau 18 Kankin 
10.0  Add 


9.10.10.10.0  5 Ahau 13 Xul 
10  Subtract 


9.10.10.9.10  8 Oc 3 Xul 
Flip 10 counters across in the uinal division, eight counters backward in the day number coefficient, and count downward 10 places in the month column. As 10 places in the month column takes one through the five nameless days—the Uayeb—one must not count this as a month but reduce the month number from 18 to 13 by flipping five counters leftward. To avoid mistakes with the Uayeb, this is placed to the left of the column of normal months. The calculation leads to 13 Xul.
To remove the 10 days by which Ahau exceeds Oc mentally add 13 to the five, the coefficient of Ahau, and then subtract 10, subtract 10 kins (as this stands at 0, “borrow” one uinal and from the resulting 20 subtract 10 just as in our arithmetic we “borrow” one of the second unit when, for example, subtracting 7 from 50). Move the day sign indicator to Oc (10 places back in the regular count), and subtract 10 from the month position. The abacus will then show 9.10.10.9.10, 8 Oc 3 Xul (fig. 4a).
Step 2. The required 8 Oc is now reached, but the month is 3 Xul,
not 13 Yax. To get the required month one uses the formula already
noted (Examples 4 and 5) that for every 20computingyear period added
(or subtracted) one subtracts (or adds) respectively one month. 13
Yax is four and onehalf months later than 3 Xul so one subtracts
five 20computingyear periods and adds five months.
9.10.10.9.10  8 Oc 3 Xul 
5.1.2.0  (364 x 20 x 5) Subtract 


9.5.9.7.10  8 Oc 3 Zac 
Flip to left (for subtraction) five, one, and two counters respectively in the katun, tun, and uinal divisions, and move the month indicator downward five places. The abacus will then register 9.5.9.7.10, 8 Oc 3 Zac (fig. 4b).
Step 3. Since our desired position was four and onehalf months
from 3 Xul, and as we were dealing in whole months, we subtracted five
20computingyear periods and overshot the position by 10 days. It
would have been equally as easy to have come 10 days short of the mark
by subtracting only four 20computingyear periods. The 10 days by
which the mark was overshot are taken care of by adding 10 computing years,
which means 10 days must be subtracted from the month position (Examples
6 and 7).
9.5.9.7.10  8 Oc 3 Zac 
10.2.0  (364 x 10) Add 


9.5.19.9.10  8 Oc 13 Yax 
Flip 10 counters to the right in the tun division and two counters in the uinal division. Take away 10 days from the month (i.e. in this case drop one month [20 days] and add 10 to the coefficient). The abacus will now register 9.5.19.9.10, 8 Oc 13 Yax, which gives the required answer that 9.5.19.9.10 is a Long Count position of 8 Oc 13 Yax (fig. 4c).
Step 4. Our novice, however, is making his calculations around
Katun 14. Inspection shows him that three Calendar Rounds should
be added to his answer. He has at hand, of course, his table for
calculating Calendar Rounds, and adds the necessary three Calendar Rounds.
9.5.19.9.10  8 Oc 3 Yax 
7.18.3.0  (3 Calendar Rounds) Add 


9.13.17.12.10  8 Oc 13 Yax 
He flips seven, 18, and three counters to the right in the katun, tun, and uinal divisions (in the tun division he completes 20 tuns, which he adds as a single counter to the katun total, and still has 17 tun counters left). The abacus now reads 9.13.17.12.10, 8 Oc, 13 Yax (fig. 4d) which is the required answer.
The two processes have taken some space to explain, but the Maya priest with constant practice could have made the calculations in little more time than is required to write them in Arabic notation. After very little practice, and without an abacus, but using clumsy cloves on a table, I attained considerable speed and reliability.



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