Blinking snake

Mayan Calendar Correlation:
A Response to Questions Raised by
Sid Hollander and Will F. McGee

Ivan Van Laningham


Subject: AS: Mayan Calendar Correlation
Date: Wed, 31 May 2000 16:18:35 -0400
From: Sid Hollander
To: AZTLAN at LISTSERV.LOUISVILLE.EDU


Ivan said:

> 4 Ahaw is tzolk’in position 159 (count from 1 Imix)

> 8 Kumk’u is haab position 348 (count from 0 Pohp)

> 4 Ahaw 8 Kumk’u is calendar round position 7283 (count from 1
> Kaban 0 Pohp)

And WF MCGee said

> Furthermore, I calculate that 04 Ahau 8 Cumku is position 2539
> after day one of the count. It (the count) could have started
> anywhere.

Sid:

I believe that 4-Ahaw is in position 160 and not 159 as you state. If you count from 1-Imix this means that the first Ahaw occurrence (7-Ahaw) will be in position 20 and since all subsequent Ahaw will be in a multiples of 20. 159 is NOT a multiple of 20.

Ivan responds:

You are correct, in a sense, in saying that 4 ’Ahaw is in “position 160.”  It is indeed the 160th day from 1 ’Imix.  You are forgetting that 1 ’Imix occupies position 0 of the tzolk’in.  As there are ten numbers that make up our base 10 (decimal) system, the digits 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9, so too are there 160 days from position 0 to position 159:  0, 1, 2, 3, 4 . and 159.  The thirteen numbers that make up the trecena go from 1 to 13, but 13 operates as if it were a mathematical 0.  The sequence can be viewed like this:

Trecena
Mathematical
Equivalent
1
1
2
2
3
3
4
4
5
5
6
6
7
7
8
8
9
9
10
10
11
11
12
12
13
0

The veintena, or twenty named days, can be viewed similarly (mathematically speaking):

Veintena
Glyph
Mathematical
Equivalent
’Ahaw
Ahaw
0
’Imix
Imix
1
’Ik
Ik
2
’Ak’bal
Akbal
3
K’an
Kan
4
Chik’chan
Chikchan
5
Kimi
Kimi
6
Manik’
Manik
7
Lamat
Lamat
8
Muluk
Muluk
9
Ok
Ok
10
Chuwen
Chuwen
11
’Eb
Eb
12
Ben
Ben
13
’Ix
Ix
14
Men
Men
15
K’ib
Kib
16
Kaban
Kaban
17
’Etz’nab
Etznab
18
Kawak
Kawak
19

Therefore, the “first” day of the trecena is numbered “one,” and the “first” day of the veintena is also numbered “one” and named “’Imix.” Modular arithmetic is that arithmetic which views all numbers as being tuples (pairs or sets) of positions in the bases involved. When we are discussing the tzolk’in, the two bases are thirteen and twenty; any given position in the 260-day cycle can be encoded as a tuple of two positions in the trecena and the veintena. (0,13), for example, represents position 0 (day 13) in the trecena and position 13 in the veintena; it translates into the tzolk’in 13 Ben. Whenever we provide a tzolk’in date, we are simultaneously giving the exact co-ordinates which will allow us to reconstruct the mathematical position in the 260-day cycle.

Q What is the position of 13 Ben in the tzolk’in?
A 12. Why?

No one seriously doubts that the Mayans started the tzolk’in on any but the day named 1 ’Imix. We refer to this as the “first day of the tzolk’in.” However, you must remember that the first position, or slot, in base 10 arithmetic is occupied by the digit zero. Even though zero represents no physical quantity, it occupies a position in decimal arithmetic, else we would be unable to describe “10” as a “two-digit number.”

Mathematically, then, the “first” position of base 260 arithmetic is occupied by (surprise) the digit zero, which would ordinarily be (0,0). The trecena starts at one, however, and the veintena begins with ’Imix, which is, mathematically, also one. In our tzolk’in notation, then, 1 ’Imix is the tuple (1,1). Therefore:

Trecena Veintena Tuple Tzolk’in position Day
1
’Imix Imix
(1,1)
0
1
2
’Ik Ik
(2,2)
1
2
3
’Ak’bal Akbal
(3,3)
2
3
4
K’an Kan
(4,4)
3
4
5
Chik’chan Chikchan
(5,5)
4
5
6
Kimi Kimi
(6,6)
5
6
7
Manik’ Manik
(7,7)
6
7
8
Lamat Lamat
(8,8)
7
8
9
Muluk Muluk
(9,9)
8
9
10
Ok Ok
(10,10)
9
10
11
Chuwen Chuwen
(11,11)
10
11
12
’Eb Eb
(12,12)
11
12
13
Ben Ben
(0,13)
12
13
4
’Ahaw Ahaw
(4,0)
159
160
12
Kawak Kawak
(12,19)
258
259
13
’Ahaw Ahaw
(0,0)
259
260

In this method the tuple notation for 4 ’Ahaw is (4,0) (look at the two tables above). The formula, according to Lounsbury, for finding the position in the tzolk’in, is this:

p=40[(tr2 - tr1) - (v2 - v1)]+(v2 - v1) % 260

(Note: the % symbol stands for the MOD, or modulo, operator.)

I’m not going to provide a detailed explanation for this formula, since I have dealt with that on my Calendar Round page. Briefly, however, the formula indicates that to find the position in the tzolk’in, you must count from a starting point. In the formula, this is 1 ’Imix, or (tr1,v1), which is, as we’ve seen above, (1,1). At heart, the formula answers the question “What is the distance from 1 ’Imix to any other day in the tzolk’in?” I would recommend following the link above and working out

  1. The distance from 1 ’Imix to 13 Ben
  2. The distance from 1 ’Imix to 13 ’Ahaw

You will find that the distance from 1 ’Imix to 13 Ben is the same as the distance from 0 to 12, 13 days; the distance from 1 ’Imix to 13 ’Ahaw is the same as the distance from 0 to 259, 260 days. If you do not believe this, count them. Write them out on paper, matching 1 ’Imix and 0, 2 ’Ik and 1, and so on. The distance from 1 ’Imix to 13 Ben is 13 days, but since we label 1 ’Imix as position 0, we must logically label 13 Ben as position 12.

In the same way, 4 ’Ahaw is the 160th day from 1 ’Imix, but because 1 ’Imix is position 0, we label 4 ’Ahaw as position 159. This is a matter of simple convenience.

Sid again:

With reference to the other two statements. Having NEVER seen ANY MAYA inscription with a 0 associated with ANY Haab month, I do not know where in the Maya Haab 0-Pohp occurs.

Ivan responds:

This statement is irrelevant. You may not have seen any inscriptions with the zero symbol associated with a haab month glyph, but you have seen the seating glyph associated with same. Mathematically, the seating glyph functions as a zero, regardless of whether it appears as “0 Pohp” or “5 Wayeb.” In either circumstance, Wayeb has five days and Pohp twenty. In the 0 Pohp case, Pohp’s days are numbered 0 through 19; in the 5 Wayeb case, Pohp has twenty days numbered 5 Wayeb through 19 Pohp.

Sid again:

Because Kumk’u is the 18 month that must mean that 17 months of 20 days (a total of 340 days) must have passed before getting to arriving at the Kumk’u. If you say that it is the first day of the month is 0, then I believe that 1-Kumk’u must be the second and 2-Kumk’u the third and 8-Kumk’u the 9th day of the month and occupy day 349 i.e. (340+9)

Ivan responds:

Yes, of course. Kumk’u, however, being the 18th month is numbered 17 (counting from month 0, which is Pohp). The position of the first day of Kumk’u, 0 Kumk’u, is found by simply multiplying 17 times 20: 340. If you add one to 340, you get 341. If you add eight days to 0 Kumk’u, you get 8 Kumk’u. 340 plus 8 is 348. Remember, however, that this is position 348, not day 348. Counting 0 Pohp as position 0 and day one, then 8 Kumk’u is indeed day 349. 4 Wayeb, the last day of the haab and by definition day 365, occupies the mathematical position, or slot, 364. The distance from 0 to 364 is 365 days, just as the distance from 1 to 365 is also 365 days.

Sid again:

Two Pet peeves:

Back to Sid:

And I will pause on the third statement for just a moment for those that like to know that 1 Kaban 0 Pohp is day 11,698 from 4-Ahaw 8-Kumk’u. I can not explain how one would get 7283 days going back in the other direction. The sum of the two should be 18,980 and not 18,981. Perhaps the 7283 should be 7282.

Ivan responds:

No, 7283 should not be 7282. The Calendar Round comprises 18,980 days; the positions range from 0 (1 Kaban 0 Pohp) to 18,979 (13 K’ib 4 Wayeb). To find the distance from 1 Kaban 0 Pohp, simply subtract. Counting forward from 1 Kaban 0 Pohp (position 0) 7,283 days gets you to position 7283 and day 7284, but always 4 ’Ahaw 8 Kumk’u. If you count forward another 11,696 days (not 11,698), you land on 13 K’ib 4 Wayeb, which is position 18,979, the last day of the Calendar Round. Again, it does not matter if you label 1 Kaban 0 Pohp as day one or position 0. What matters is that if you add the size of the Calendar Round, 18,980 days, to any given (legal) Mayan date, you end up on the same day you started from. If you start on 1 Kaban 0 Pohp and add 18,980 days, you will end up on 1 Kaban 0 Pohp.  The interval, or distance, from 0 to 7283 is the same as that from 1 to 7284.  The distance from 7283 to 18979 is identical to the distance from 7284 to 18980.

0.0. 0.0. 0  4 ’Ahaw 8 Kumk’u (CR position 7283)

0.1.12.8.16  (11696) add

0.1.12.8.16  13 K’ib 4 Wayeb (CR position 18979)


 0.0.0.0.0  4 ’Ahaw 8 Kumk’u (CR position 7283)

-0.1.0.4.3  (7283) subtract

-0.1.0.4.3  1 Kaban 0 Pohp (CR position 0)


0.1.12.8.16  13 K’ib 4 Wayeb (CR position 18979)

0.0. 0.0. 1  (1) add

0.1.12.8.17  1 Kaban 0 Pohp (CR position 0)


0.0. 0. 0.0  4 ’Ahaw 8 Kumk’u (CR position 7283)

0.2.12.13.0  (11980) add

0.2.12.13.0  4 ’Ahaw  8 Kumk’u (CR position 7283)

I think it is clear from the foregoing discussion that the conventional starting date of the Mayan Calendar, 4 ’Ahaw 8 Kumk’u, is not the same as the starting date of the Calendar Round, or the tzolk’in, or the haab, or the 819-day cycle, and in fact, not of anything except the Long Count.


Will F. McGee writes:

Dear Ivan,

I most respectfully disagree with your analysis of 04 Ahau 8 Cumku as position 159. The reason is that the Haab kin number count is regular because 365 is 1 MOD 13. That is to say, 364 is evenly divisible by 13. Therefore, if the Haab begins with the kin number 1 it will also end on kin number 1 — begin 2 end 2 — begin 3 end 3 etc.

Ivan responds:

I assume that “position 159” is a typo and that you meant “position 7283” for 4 ’Ahaw 8 Kumk’u.

The number of days in the haab is 365, and 365 % 13 = 1 (as you point out). This means that the trecena increments by one day for each haab incremented. 365 % 20 = 5, and therefore the veintena increments by 5 days for each haab incremented. The haab position remains the same:

0.0.0.0.0  4 ’Ahaw 8 Kumk’u (CR position 7283)

0.0.1.0.5  (365) add

0.0.1.0.5  5 Chik’chan 8 Kumk’u (CR position 7648)

Note that ’Ahaw is treated mathematially as zero; Chik’chan is day 5 in the veintena (which of course must match the k’in position of the Long Count). It is true that if a haab begins on a particular trecena day it will also end on that same day; that is, if a haab begins on 8 Kaban 0 Pohp, it must end on day 8 of the trecena. This means that the veintena only increments by 4 days, however.

0.0.0.0.17  8 Kaban 0 Pohp (CR position 7300)

0.0.1.0.4   (364) add

0.0.1.1.1   8 ’Imix 4 Wayeb (CR position 7664)

Will:

At 04 Ahau 8 Cumku the next day must be 05 Imix 9 Cumku etc. Therefore, 04 Ahau 8 Cumku + 16 is the end of that particular 365 Haab. That day is kin number 7.

Ivan:

0.0.0.0.0   4 ’Ahaw 8 Kumk’u (CR position 7283)

0.0.0.0.16  add

0.0.0.0.16  7 K’ib 4 Wayeb (CR position 7299)

Will:

Therefore, 7 x 365 = 2555 and 2555 - 16 = 2539. This day is 4 MOD 13.

Check me out.

Ivan responds:

You’re on the right track here. As you pointed out above, each haab that passes increments the trecena day by one, and thus, there ought to be a way to recover, from a complete Calendar Round date, the total number of haabs that have passed since the beginning of the Calendar Round. The trecena day number only gets us a thirteen-haab cycle.  The veintena, which increments by five days on each haab, lets us have only four legal veintena days at the beginning of any given haab. These are ’Ik, Manik’, ’Eb and Kaban. Thus, the Calendar Round is constrained to begin on a day 0 Pohp with a veintena of one of these four days.

OK, let’s plug in your numbers. We’re trying to find the beginning of the Calendar Round, so let’s work from the beginning of the very first haab after 4 ’Ahaw 8 Kumk’u:

 0.0.0.0.17  8 Kaban 0 Pohp (CR position 7300)

 0.0.7.1.15  (7 · 365) subtract

-0.0.7.0.18  1 ’Ik’ 0 Pohp (CR position 4745)

This does indeed put us on a day 0 Pohp and on a legal veintena day (’Ik’), and it is, as you say, 2555 days before the beginning of the first haab after the zero date, 4 ’Ahaw 8 Kumk’u. However, 1 ’Ik’ is position 221 in the tzolk’in, which is further from the starting point. Notice that no matter how we determine the starting point for the CR, we must allow for the existence of the particular combination of 4 ’Ahaw 8 Kumk’u. 0 Pohp is the beginning of the haab; if the tzolk’in were aligned relative to the haab so that position 0 of both began on the same day, we would have a beginning point of 1 ’Imix 0 Pohp; but if this were the case we could never have 4 ’Ahau 8 Kumk’u. It would be an illegal combination. Notice also that given the existing allowable combinations, ’Imix is not a legal day on which the Calendar Round can begin.

We must, therefore, follow a different approach. Notice that the 260-day tzolk’in and the 365-day haab are locked into a particular relation which can be deduced by working from the 4 ’Ahaw 8 Kumk’u date. We know that 4 ’Ahaw is position 159 in the tzolk’in and that 8 Kumk’u is position 348 in the haab. Notice also that the length of the Calendar Round is 18,980 days, and that this is 73 times 260, or 52 times 365. If 260 and 365 had no lowest common denominator other than 1, the CR would be 260 · 365, or 94,900 days long. Since the lowest common denominator is 5, however, we can simply divide the 94,900 by 5, and discover that the answer is 18,980—exactly what we’ve been told all along.

Floyd Lounsbury has pointed out that we can, indeed, find out how many haabs have passed since the beginning (or from any arbitrary point) of the Calendar Round by examining the difference between the tzolk’in and haab positions. The formula is

nH=(tz - h) % 52

Again, I’m not going to provide a detailed explanation of this formula because it’s covered on my Calendar Round page.  We can plug in our values from above, 159 and 348:

nH=(159 - 348) % 52
nH=-189 % 52
nH=19

Which indicates that the minimum distance between 4 ’Ahaw 8 Kumk’u and the beginning of the Calendar Round is 19 whole haabs, or 6,935 days. Let’s subtract 19 haabs from the zero date:

 0.0. 0.0.0   4 ’Ahaw 8 Kumk’u (CR position 7283)

 0.0.19.4.15  (19 · 365) subtract

-0.0.19.4.15  11 Chik’chan 8 Kumk’u (CR position 348)

Obviously, CR position 348 is not the beginning of the Calendar Round, since we know we must begin on a 0 Pohp, and 11 Chik’chan is position 244 in the tzolk’in. We need to take into account the haab position for 4 ’Ahau 8 Kumk’u (348), and that is covered by another formula from Lounsbury:

pCR=365 · nH + h

As before, we can plug our values into the formula:

pCR=365 · 19 + 348
pCR=6935 + 348
pCR=7283

If we then subtract 7283 from our zero date, we find:

 0.0.0.0.0   4 ’Ahaw 8 Kumk’u (CR position 7283)

 0.1.0.4.3   (19 · 365 + 348) subtract

-0.1.0.4.3   1 Kaban 0 Pohp (CR position 0)

The tzolk’in position of 1 Kaban 0 Pohp, applying the formula p=40[(tr2 - tr1) - (v2 - v1)]+(v2 - v1) % 260, is 156 counting from position 0, 1 ’Imix. It is also haab position 0; 0 Pohp is the beginning of the haab, or day 0. Thus, the Calendar Round position (pCR) for 4 ’Ahaw 8 Kumk’u is 7283.


References


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