Chapter 14 of 31 · 11569 words · ~58 min read

Chapter VI

The numbers from 21 to 359, inclusive, involved the use of two places--the kin place and the uinal place--which, according to Table VIII, we saw had numerical values of 1 and 20, respectively. For example, the number 37 was expressed as shown in figure 63, b. The 17 in the kin place has a value of 17 (17 × 1) and the 1 in the uinal, or second, place a value of 20 (1 (the numeral) × 20 (the fixed numerical value of the second place)). The sum of these two products equals 37. Again, 300 was written as in figure 63, c. The 0 in the kin place has the value 0 (0 × 1), and the 15 in the second place has the value of 300 (15 × 20), and the sum of these products equals 300.

To express the numbers 360 to 7,199, inclusive, three places or terms were necessary--kins, uinals, and tuns--of which the last had a numerical value of 360. (See Table VIII.) For example, the number 360 is shown in figure 63, d. The 0 in the lowest place indicates that 0 kins are involved, the 0 in the second place indicates that 0 uinals or 20's are involved, while the 1 in the third place shows that there is 1 tun, or 360, kins recorded (1 (the numeral) × 360 (the fixed numerical value of the third position)); the sum of these three products equals 360. Again, the number 7,113 is expressed as shown in figure 63, e. {131} The 13 in the lowest place equals 13 (13 × 1); the 13 in the second place, 260 (13 × 20); and the 19 in the third place, 6,840 (19 × 360). The sum of these three products equals 7,113 (13 + 260 + 6,840),

[Illustration: FIG. 63. Examples of the second method of numeration, used exclusively in the codices.]

The numbers from 7,200 to 143,999, inclusive, involved the use of four places or terms--kins, uinals, tuns, and katuns--the last of which (the fourth place) had a numerical value of 7,200. (See Table VIII.) For example, the number 7,202 is recorded in figure 63, _f_. {132} The 2 in the first place equals 2 (2×1); the 0 in the second place, 0 (0×20); the 0 in the third place, 0 (0×360); and the 1 in the fourth place, 7,200 (1×7,200). The sum of these four products equals 7,202 (2+0+0+7,200). Again, the number 100,932 is recorded in figure 63, _g_. Here the 12 in the first place equals 12 (12×1); the 6 in the second place, 120 (6×20); the 0 in the third place, 0 (0×360); and the 14 in the fourth place, 100,800 (14×7,200). The sum of these four products equals 100,932 (12+120+0+100,800).

The numbers from 144,000 to 2,879,999, inclusive, involved the use of five places or terms--kins, uinals, tuns, katuns, and cycles. The last of these (the fifth place) had a numerical value of 144,000. (See Table VIII.) For example, the number 169,200 is recorded in figure 63, _h_. The 0 in the first place equals 0 (0×1); the 0 in the second place, 0 (0×20); the 10 in the third place, 3,600 (10×360); the 3 in the fourth place, 21,600 (3×7,200); and the 1 in the fifth place, 144,000 (1×144,000). The sum of these five products equals 169,200 (0+0+3,600+21,600+144,000). Again, the number 2,577,301 is recorded in figure 63, _i_. The 1 in the first place equals 1 (1×1); the 3 in the second place, 60 (3×20); the 19 in the third place, 6,840 (19×360); the 17 in the fourth place, 122,400 (17×7,200); and the 17 in the fifth place, 2,448,000 (17x144,000). The sum of these five products equals 2,577,301 (1+60+6,480+122,400+2,448,000).

The writing of numbers above 2,880,000 up to and including 12,489,781 (the highest number found in the codices) involves the use of six places, or terms--kins, uinals, tuns, katuns, cycles, and great cycles--the last of which (the sixth place) has the numerical value 2,880,000. It will be remembered that some have held that the sixth place in the inscriptions contained only 13 units of the fifth place, or 1,872,000 units of the first place. In the codices, however, there are numerous calendric checks which prove conclusively that in so far as the codices are concerned the sixth place was composed of 20 units of the fifth place. For example, the number 5,832,060 is expressed as in figure 63, _j_. The 0 in the first place equals 0 (0×1); the 3 in the second place, 60 (3×20); the 0 in the third place, 0 (0×360); the 10 in the fourth place, 72,000 (10×7,200); the 0 in the fifth place, 0 (0×144,000); and the 2 in the sixth place, 5,760,000 (2×2,880,000). The sum of these six terms equals 5,832,060 (0+60+0+72,000+0+5,760,000). The highest number in the codices, as explained above, is 12,489,781, which is recorded on page 61 of the Dresden Codex. This number is expressed as in figure 63, _k_. The 1 in the first place equals 1 (1×1); the 15 in the second place, 300 (15×20); the 13 in the third place, 4,680 (13×360); the 14 in the fourth place, 100,800 (14×7,200); the 6 in the fifth place, 864,000 (6×144,000); and the 4 in the sixth place, 11,520,000 (4×2,880,000). The sum of these six products equals 12,489,781 (1+300+4,680+100,800+864,000+11,520,000). {133}

It is clear that in numeration by position the order of the units could not be reversed as in the first method without seriously affecting their numerical values. This must be true, since in the second method the numerical values of the numerals depend entirely on their position--that is, on their distance above the bottom or first term. In the first method, the multiplicands--the period glyphs, each of which had a fixed numerical value--are always expressed[86] with their corresponding multipliers--the numerals 0 to 19, inclusive; in other words, the period glyphs themselves show whether the series is an ascending or a descending one. But in the second method the multiplicands are not expressed. Consequently, since there is nothing about a column of bar and dot numerals which in itself indicates whether the series is an ascending or a descending one, and since in numeration by position a fixed starting point is absolutely essential, in their second method the Maya were obliged not only to fix arbitrarily the direction of reading, as from bottom to top, but also to confine themselves exclusively to the presentation of one kind of series only--that is, ascending series. Only by means of these two arbitrary rules was confusion obviated in numeration by position.

However dissimilar these two methods of representing the numbers may appear at first sight, fundamentally they are the same, since both have as their basis the same vigesimal system of numeration. Indeed, it can not be too strongly emphasized that throughout the range of the Maya writings, codices, inscriptions, or Books of Chilam Balam[87] the several methods of counting time and recording events found in each are all derived from the same source, and all are expressions of the same numerical system.

That the student may better grasp the points of difference between the two methods they are here contrasted:

TABLE XII. COMPARISON OF THE TWO METHODS OF NUMERATION

We have seen in the foregoing pages (1) how the Maya wrote their 20 {134} numerals, and (2) how these numerals were used to express the higher numbers. The next question which concerns us is, How did they use these numbers in their calculations; or in other words, how was their arithmetic applied to their calendar? It may be said at the very outset in answer to this question, that in so far as known, _numbers appear to have had but one use throughout the Maya texts, namely, to express the time elapsing between dates_.[88] In the codices and the inscriptions alike all the numbers whose use is understood have been found to deal exclusively with the counting of time.

This highly specialized use of the numbers in Maya texts has determined the first step to be taken in the process of deciphering them. Since the primary unit of the calendar was the day, all numbers should be reduced to terms of this unit, or in other words, to units of the first order, or place.[89] Hence, we may accept the following as the _first step_ in ascertaining the meaning of any number:

FIRST STEP IN SOLVING MAYA NUMBERS

Reduce all the units of the higher orders to units of its first, or lowest, order, and then add the resulting quantities together.

The application of this rule to any Maya number, no matter of how many terms, will always give the actual number of primary units which it contains, and in this form it can be more conveniently utilized in connection with the calendar than if it were left as recorded, that is, in terms of its higher orders.

The reduction of units of the higher orders to units of the first order has been explained on pages 105-133, but in order to provide the student with this same information in a more condensed and accessible form, it is presented in the following tables, of which Table XIII is to be used for reducing numbers to their primary units in the inscriptions, and Table XIV for the same purpose in the codices. {135}

TABLE XIII. VALUES OF HIGHER PERIODS IN TERMS OF LOWEST, IN INSCRIPTIONS

1 great cycle = [90]2,880,000 1 cycle 144,000 1 katun 7,200 1 tun 360 1 uinal 20 1 kin 1

TABLE XIV. VALUES OF HIGHER PERIODS IN TERMS OF LOWEST, IN CODICES

1 unit of the 6th place = 2,880,000 1 unit of the 5th place 144,000 1 unit of the 4th place 7,200 1 unit of the 3d place 360 1 unit of the 2d place 20 1 unit of the 1st place 1

It should be remembered, in using these tables, that each of the signs for the periods therein given has its own particular numerical value, and that this value in each case is a multiplicand which is to be multiplied by the numeral attached to it (not shown in Table XIII). For example, a 3 attached to the katun sign reduces to 21,600 units of the first order (3×7,200). Again, 5 attached to the uinal sign reduces to 100 units of the first order (5×20). In using Table XIV, however, it should be remembered that the position of a numeral multiplier determines at the same time that multiplier's multiplicand. Thus a 5 in the third place indicates that the 5's multiplicand is 360, the numerical value of the third place, and such a term reduces to 1,800 units of the first place (5×360=1,800). Again, a 10 in the fourth place indicates that the 10's multiplicand is 7,200, the numerical value corresponding to the fourth place, and such a term reduces to 72,000 units of the first place.

Having reduced all the terms of a number to units of the 1st order, the next step in finding out its meaning is to discover the date from which it is counted. This operation gives rise to the _second step_.

SECOND STEP IN SOLVING MAYA NUMBERS

Find the date from which the number is counted:

This is not always an easy matter, since the dates from which Maya numbers are counted are frequently not expressed in the texts; consequently, it is clear that no single rule can be formulated which will cover all cases. There are, however, two general rules which will be found to apply to the great majority of numbers in the texts:

_Rule 1._ When the starting point or date is expressed, usually, though not invariably, it precedes[91] the number counted from it.

It should be noted, however, in connection with this rule, that the starting date hardly ever immediately precedes the number from which it is counted, but that several glyphs nearly always stand {136} between.[92] Certain exceptions to the above rule are by no means rare, and the student must be continually on the lookout for such reversals of the regular order. These exceptions are cases in which the starting date (1) follows the number counted from it, and (2) stands elsewhere in the text, entirely disassociated from, and unattached to, the number counted from it.

The second of the above-mentioned general rules, covering the majority of cases, follows:

_Rule 2_. When the starting point or date is not expressed, if the number is an Initial Series the date from which it should be counted will be found to be 4 Ahau 8 Cumhu.[93]

This rule is particularly useful in deciphering numbers in the inscriptions. For example, when the student finds a number which he can identify as an Initial Series,[94] he may assume at once that such a number in all probability is counted from the date 4 Ahau 8 Cumhu, and proceed on this assumption. The exceptions to this rule, that is, cases in which the starting point is not expressed and the number is not an Initial Series, are not numerous. No rule can be given covering all such cases, and the starting points of such numbers can be determined only by means of the calculations given under the third and fourth steps, below.

Having determined the starting point or date from which a given number is to be counted (if this is possible), the next step is to find out which way the count runs; that is, whether it is _forward_ from the starting point to some _later date_, or whether it is _backward_ from the starting point to some _earlier date_. This process may be called the _third step_.

THIRD STEP IN SOLVING MAYA NUMBERS

Ascertain whether the number is to be counted forward or backward from its starting point.

It may be said at the very outset in this connection that the overwhelming majority of Maya numbers are counted _forward_ from their starting points and not backward. In other words, they proceed from _earlier to later dates_ and not vice versa. Indeed, the preponderance of the former is so great, and the exceptions are so rare, that the student should always proceed on the postulate that the count is forward until proved definitely to be otherwise. {137}

[Illustration: FIG. 64. Figure showing the use of the "minus" or "backward" sign in the codices.]

In the codices, moreover, when the count is backward, or contrary to the general practice, the fact is clearly indicated[95] by a special character. This character, although attached only to the lowest term[96] of the number which is to be counted backward, is to be interpreted as applying to all the other terms as well, its effect extending to the number as a whole. This "backward sign" (shown in fig. 64) is a circle drawn in red around the lowest term of the number which it affects, and is surmounted by a knot of the same color. An example covering the use of this sign is given in figure 64. Although the "backward sign" in this figure surrounds only the numeral in the first place, 0, it is to be interpreted, as we have seen, as applying to the 2 in the second place and the 6 in the third place. This number, expressed as 6 tuns, 2 uinals, and 0 kins, reduces to 2,200 units of the first place, and in this form may be more readily handled (first step). Since the starting point usually precedes the number counted from it and since in figure 64 the number is expressed by the second method, its starting point will be found standing below it. This follows from the fact that in numeration by position the order is from bottom to top. Therefore the starting point from which the 2,200 recorded in figure 64 is counted will be found to be below it, that is, the date 4 Ahau 8 Cumhu[97] (second step). Finally, the red circle and knot surrounding the lowest (0) term of this 2,200 indicates that this number is to be counted _backward_ from its starting point, not forward (third step).

On the other hand, in the inscriptions no special character seems to have been used with a number to indicate that it was to be counted backward; at least no such sign has yet been discovered. In the inscriptions, therefore, with the single exception[98] mentioned below, the student can only apply the general rule given on page 136, that in the great majority of cases the count is forward. This rule will be found to apply to at least nine out of every ten numbers. The exception above noted, that is, where the practice is so uniform as to render possible the formulation of an unfailing rule, has to do with Initial Series. This rule, to which there are no known exceptions, may be stated as follows:

_Rule 1_. In Initial Series the count is _always forward_, and, in general throughout the inscriptions. The very few cases in which the count _is_ backward, are confined chiefly to Secondary Series, and it is in {138} dealing with this kind of series that the student will find the greatest number of exceptions to the general rule.

Having determined the direction of the count, whether it is forward or backward, the next (_fourth_) step may be given.

FOURTH STEP IN SOLVING MAYA NUMBERS

To count the number from its starting point.

We have come now to a step that involves the consideration of actual arithmetical processes, which it is thought can be set forth much more clearly by the use of specific examples than by the statement of general rules. Hence, we will formulate our rules after the processes which they govern have been fully explained.

In counting any number, as 31,741, or 4.8.3.1 as it would be expressed in Maya notation,[99] from any date, as 4 Ahau 8 Cumhu, there are four unknown elements which have to be determined before we can write the date which the count reaches. These are:

1. The day coefficient, which must be one of the numerals 1 to 13, inclusive.

2. The day name, which must be one of the twenty given in Table I.

3. The position of the day in some division of the year, which must be one of the numerals 0 to 19, inclusive.

4. The name of the division of the year, which must be one of the nineteen given in Table III.

These four unknown elements all have to be determined from (1) the starting date, and (2) the number which is to be counted from it.

If the student will constantly bear in mind that all Maya sequences, whether the day coefficients, day signs, positions in the divisions of the year, or what not, are absolutely continuous, repeating themselves without any break or interruption whatsoever, he will better understand the calculations which follow.

It was explained in the text (see pp. 41-44) and also shown graphically in the tonalamatl wheel (pl. 5) that after the day coefficients had reached the number 13 they returned to 1, following each other indefinitely in this order without interruption. It is clear, therefore, that the highest multiple of 13 which the given number contains may be subtracted from it without affecting in any way the value of the day coefficient of the date which the number will reach when counted from the starting point. This is true, because no matter what the day coefficient of the starting point may be, any multiple of 13 will always bring the count back to the same day coefficient. {139}

Taking up the number, 31,741, which we have chosen for our first example, let us deduct from it the highest multiple of 13 which it contains. This will be found by dividing the number by 13, and multiplying the _whole-number part_ of the resulting quotient by 13: 31,741 ÷ 13 = 2,441-8/13. Multiplying 2,441 by 13, we have 31,733, which is the highest multiple of 13 that 31,741 contains; consequently it may be deducted from 31,741 without affecting the value of the resulting day coefficient: 31,741 - 31,733 = 8. In the example under consideration, therefore, 8 is the number which, if counted from the day coefficient of the starting point, will give the day coefficient of the resulting date. In other words, after dividing by 13 the only part of the resulting quotient which is used in determining the new day coefficient is the _numerator_ of the fractional part.[100] Hence the following rule for determining the first unknown on page 138 (the day coefficient):

_Rule 1._ To find the new day coefficient divide the given number by 13, and count forward the numerator of the fractional part of the resulting quotient from the starting point if the count is forward, and backward if the count is backward, deducting 13 in either case from the resulting number if it should exceed 13.

Applying this rule to 31,741, we have seen above that its division by 13 gives as the fractional part of the quotient 8/13. Assuming that the count is forward from the starting point, 4 Ahau 8 Cumhu, if 8 (the numerator of the fractional part of the quotient) be counted forward from 4, the day coefficient of the starting point (4 Ahau 8 Cumhu), the day coefficient of the resulting date will be 12 (4 + 8). Since this number is below 13, the last sentence of the above rule has no application in this case. In counting forward 31,741 from the date 4 Ahau 8 Cumhu, therefore, the day coefficient of the resulting date will be 12; thus we have determined our first unknown. Let us next find the second unknown, the day sign to which this 12 is prefixed.

It was explained on page 37 that the twenty day signs given in Table I succeed one another in endless rotation, the first following immediately the twentieth no matter which one of the twenty was chosen as the first. Consequently, it is clear that the highest multiple of 20 which the given number contains may be deducted from it without affecting in any way the name of the day sign of the date which the number will reach when counted from the starting point. This is true because, no matter what the day sign of the starting point may be, any multiple of 20 will always bring the count back to the same day sign. {140}

Returning to the number 31,741, let us deduct from it the highest multiple of 20 which it contains, found by dividing the number by 20 and multiplying the whole number part of the resulting quotient by 20; 31,741 ÷ 20 = 1,587-1/20. Multiplying 1,587 by 20, we have 31,740, which is the highest multiple of 20 that 31,741 contains, and which may be deducted from 31,741 without affecting the resulting day sign; 31,741 - 31,740 = 1. Therefore in the present example 1 is the number which, if counted forward from the day sign of the starting point in the sequence of the 20 day signs given in Table I, will reach the day sign of the resulting date. In other words, after dividing by 20 the only part of the resulting quotient which is used in determining the new day sign is the numerator of the fractional part. Thus we may formulate the rule for determining the second unknown on page 138 (the day sign):

_Rule 2._ To find the new day sign, divide the given number by 20, and count forward the numerator of the fractional part of the resulting quotient from the starting point in the sequence of the twenty day signs given in Table I, if the count is forward, and backward if the count is backward, and the sign reached will be the new day sign.

Applying this rule to 31,741, we have seen above that its division by 20 gives us as the fractional part of the quotient, 1/20. Since the count was forward from the starting point, if 1 (the numerator of the fractional part of the quotient) be counted forward in the sequence of the 20 day signs in Table I from the day sign of the starting point, Ahau (4 Ahau 8 Cumhu), the day sign reached will be the day sign of the resulting date. Counting forward 1 from Ahau in Table I, the day sign Imix is reached, and Imix, therefore, will be the new day sign. Thus our second unknown is determined.

By combining the above two values, the 12 for the first unknown and Imix for the second, we can now say that in counting forward 31,741 from the date 4 Ahau 8 Cumhu, the day reached will be 12 Imix. It remains to find what position this particular day occupied in the 365-day year, or haab, and thus to determine the third and fourth unknowns on page 138. Both of these may be found at one time by the same operation.

It was explained on pages 44-51 that the Maya year, at least in so far as the calendar was concerned, contained only 365 days, divided into 18 uinals of 20 days each, and the _xma kaba kin_ of 5 days; and further, that when the last position in the last division of the year (4 Uayeb) was reached, it was followed without interruption by the first position of the first division of the next year (0 Pop); and, finally, that this sequence was continued indefinitely. Consequently it is clear that the highest multiple of 365 which the given number contains may be subtracted from it without affecting in any way the position in the year of the day which the number will reach when {141} counted from the starting point. This is true, because no matter what position in the year the day of the starting point may occupy, any multiple of 365 will bring the count back again to the same position in the year.

Returning again to the number 31,741, let us deduct from it the highest multiple of 365 which it contains. This will be found by dividing the number by 365 and multiplying the whole number part of the resulting quotient by 365: 31,741 ÷ 365 = 86-351/365. Multiplying 86 by 365, we have 31,390, which is the highest multiple that 31,741 contains. Hence it may be deducted from 31,741 without affecting the position in the year of the resulting day; 31,741 - 31,390 = 351. Therefore, in the present example, 351 is the number which, if counted forward from the year position of the starting date in the sequence of the 365 positions in the year, given in Table XV, will reach the position in the year of the day of the resulting date. This enables us to formulate the rule for determining the third and fourth unknowns on page 138 (the position in the year of the day of the resulting date):

_Rule 3._ To find the position in the year of the new day, divide the given number by 365 and count forward the numerator of the fractional part of the resulting quotient from the year position of the starting point in the sequence of the 365 positions of the year shown in Table XV, if the count is forward; and backward if the count is backward, and the position reached will be the position in the year which the day of the resulting date will occupy.

TABLE XV. THE 365 POSITIONS IN THE MAYA YEAR

+-----+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+ |Month|P |U |Z |Z |T |X |Y |M |C |Y |Z |C |M |K |M |P |K |C |U | | |o |o |i |o |z |u |a |o |h |a |a |e |a |a |u |a |a |u |a | | |p | |p |t |e |l |x |l |e |x |c |h |c |n |a |x |y |m |y | | | | | |z |c | |k | |n | | | | |k |n | |a |h |e | | | | | | | | |i | | | | | | |i | | |b |u |b | | | | | | | | |n | | | | | | |n | | | | | | +-----+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+ |Posi-| | | | | | | | | | | | | | | | | | | | |tion |0 |0 |0 |0 |0 |0 |0 |0 |0 |0 |0 |0 |0 |0 |0 |0 |0 |0 |0 | +-----+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+ | Do |1 |1 |1 |1 |1 |1 |1 |1 |1 |1 |1 |1 |1 |1 |1 |1 |1 |1 |1 | +-----+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+ | Do |2 |2 |2 |2 |2 |2 |2 |2 |2 |2 |2 |2 |2 |2 |2 |2 |2 |2 |2 | +-----+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+ | Do |3 |3 |3 |3 |3 |3 |3 |3 |3 |3 |3 |3 |3 |3 |3 |3 |3 |3 |3 | +-----+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+ | Do |4 |4 |4 |4 |4 |4 |4 |4 |4 |4 |4 |4 |4 |4 |4 |4 |4 |4 |4 | +-----+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+ | Do |5 |5 |5 |5 |5 |5 |5 |5 |5 |5 |5 |5 |5 |5 |5 |5 |5 |5 |--| +-----+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+ | Do |6 |6 |6 |6 |6 |6 |6 |6 |6 |6 |6 |6 |6 |6 |6 |6 |6 |6 |--| +-----+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+ | Do |7 |7 |7 |7 |7 |7 |7 |7 |7 |7 |7 |7 |7 |7 |7 |7 |7 |7 |--| +-----+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+ | Do |8 |8 |8 |8 |8 |8 |8 |8 |8 |8 |8 |8 |8 |8 |8 |8 |8 |8 |--| +-----+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+ | Do |9 |9 |9 |9 |9 |9 |9 |9 |9 |9 |9 |9 |9 |9 |9 |9 |9 |9 |--| +-----+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+ | Do |10|10|10|10|10|10|10|10|10|10|10|10|10|10|10|10|10|10|--| +-----+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+ | Do |11|11|11|11|11|11|11|11|11|11|11|11|11|11|11|11|11|11|--| +-----+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+ | Do |12|12|12|12|12|12|12|12|12|12|12|12|12|12|12|12|12|12|--| +-----+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+ | Do |13|13|13|13|13|13|13|13|13|13|13|13|13|13|13|13|13|13|--| +-----+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+ | Do |14|14|14|14|14|14|14|14|14|14|14|14|14|14|14|14|14|14|--| +-----+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+ | Do |15|15|15|15|15|15|15|15|15|15|15|15|15|15|15|15|15|15|--| +-----+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+ | Do |16|16|16|16|16|16|16|16|16|16|16|16|16|16|16|16|16|16|--| +-----+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+ | Do |17|17|17|17|17|17|17|17|17|17|17|17|17|17|17|17|17|17|--| +-----+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+ | Do |18|18|18|18|18|18|18|18|18|18|18|18|18|18|18|18|18|18|--| +-----+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+ | Do |19|19|19|19|19|19|19|19|19|19|19|19|19|19|19|19|19|19|--| +-----+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+

{142}

Applying this rule to the number 31,741, we have seen above that its division by 365 gives 351 as the numerator of the fractional part of its quotient. Assuming that the count is forward from the starting point, it will be necessary, therefore, to count 351 forward in Table XV from the position 8 Cumhu, the position of the day of the starting point, 4 Ahau 8 Cumhu.

A glance at the month of Cumhu in Table XV shows that after the position 8 Cumhu there are 11 positions in that month; adding to these the 5 in Uayeb, the last division of the year, there will be in all 16 more positions before the first of the next year. Subtracting these from 351, the total number to be counted forward, there remains the number 335 (351-16), which must be counted forward in Table XV from the beginning of the year. Since each of the months has 20 positions, it is clear that 16 months will be used before the month is reached in which will fall the 335th position from the beginning of the year. In other words, 320 positions of our 335 will exactly use up all the positions of the first 16 months, namely, Pop, Uo, Zip, Zotz, Tzec, Xul, Yaxkin, Mol, Chen, Yax, Zac, Ceh, Mac, Kankin, Muan, Pax, and will bring us to the beginning of the 17th month (Kayab) with still 15 more positions to count forward. If the student will refer to this month in Table XV he will see that 15 positions counted forward in this month will reach the position 14 Kayab, which is also the position reached by counting forward 31,741 positions from the starting position 8 Cumhu.

Having determined values for all of the unknowns on page 138, we can now say that if the number 31,741 be counted forward from the date 4 Ahau 8 Cumhu, the date 12 Imix 14 Kayab will be reached. To this latter date, i. e., the date reached by any count, the name "terminal date" has been given. The rules indicating the processes by means of which this terminal date is reached apply also to examples where the count is _backward_, not forward, from the starting point. In such cases, as the rules say, the only difference is that the numerators of the fractional parts of the quotients resulting from the different divisions are to be counted backward from the starting points, instead of forward as in the example above given.

Before proceeding to apply the rules by means of which our fourth step or process (see p. 138) may be carried out, a modification may sometimes be introduced which will considerably decrease the size of the number to be counted without affecting the values of the several parts of its resulting terminal date.

We have seen on pages 51-60 that in Maya chronology there were possible only 18,980 different dates--that is, combinations of the 260 days and the 365 positions of the year--and further, that any given day of the 260 could return to any given position of the 365 only after the lapse of 18,980 days, or 52 years. {143}

Since the foregoing is true, it follows, that this number 18,980 or any multiple thereof, may be deducted from the number which is to be counted without affecting in any way the terminal date which the number will reach when counted from the starting point. It is obvious that this modification applies only to numbers which are above 18,980, all others being divided by 13, 20, and 365 directly, as indicated in rules 1, 2, and 3, respectively. This enables us to formulate another rule, which should be applied to the number to be counted before proceeding with rules 1, 2, and 3 above, if that number is above 18,980.

_Rule_. If the number to be counted is above 18,980, first deduct from it the highest multiple of 18,980 which it contains.

This rule should be applied whenever possible, since it reduces the size of the number to be handled, and consequently involves fewer calculations.

In Table XVI are given 80 Calendar Rounds, that is, 80 multiples of 18,980, in terms of both the Maya notation and our own. These will be found sufficient to cover most numbers.

Applying the above rule to the number 31,741, which was selected for our first example, it is seen by Table XVI that 1 Calendar Round, or 18,980 days, may be deducted from it; 31,741 - 18,980 = 12,761. In other words, we can count the number 12,761 forward (or backward had the count been backward in our example) from the starting point 4 Ahau 8 Cumhu, and reach exactly the same terminal date as though we had counted forward 31,741, as in the first case.

Mathematical proof of this point follows:

12,761 ÷ 13 = 981-8/13 12,761 ÷ 20 = 638-1/20 12,761 ÷ 365 = 34-351/365

The numerators of the fractions in these three quotients are 8, 1, and 351; these are identical with the numerators of the fractions in the quotients obtained by dividing 31,741 by the same divisors, those indicated in rules 1, 2, and 3, respectively. Consequently, if these three numerators be counted forward from the corresponding parts of the starting point, 4 Ahau 8 Cumhu, the resulting terms together will form the corresponding parts of the same terminal date, 12 Imix 14 Kayab.

Similarly it could be shown that 50,721 or 69,701 counted forward or backward from any starting point would both reach this same terminal date, since subtracting 2 Calendar Rounds, 37,960 (see Table XVI), from the first, and 3 Calendar Rounds, 56,940 (see Table XVI), from the second, there would remain in each case 12,761. The student will find his calculations greatly facilitated if he will apply this rule whenever possible. To familiarize the student with the working of these rules, it is thought best to give several additional examples involving their use. {144}

TABLE XVI. 80 CALENDAR ROUNDS EXPRESSED IN ARABIC AND MAYA NOTATION

+--------+---------+-----------------+ |Calendar| Days| Cycles, Etc.| | Rounds| | | +--------+---------+-----------------+ | 1| 18,980| 2. 12. 13. 0| +--------+---------+-----------------+ | 2| 37,960| 5. 5. 8. 0| +--------+---------+-----------------+ | 3| 56,940| 7. 18. 3. 0| +--------+---------+-----------------+ | 4| 75,920| 10. 10. 16. 0| +--------+---------+-----------------+ | 5| 94,900| 13. 3. 11. 0| +--------+---------+-----------------+ | 6| 113,880| 15. 16. 6. 0| +--------+---------+-----------------+ | 7| 132,860| 18. 9. 1. 0| +--------+---------+-----------------+ | 8| 151,840| 1. 1. 1. 14. 0| +--------+---------+-----------------+ | 9| 170,820| 1. 3. 14. 9. 0| +--------+---------+-----------------+ | 10| 189,800| 1. 6. 7. 4. 0| +--------+---------+-----------------+ | 11| 208,780| 1. 8. 19. 17. 0| +--------+---------+-----------------+ | 12| 227,760| 1. 11. 12. 12. 0| +--------+---------+-----------------+ | 13| 246,740| 1. 14. 5. 7. 0| +--------+---------+-----------------+ | 14| 265,720| 1. 16. 18. 2. 0| +--------+---------+-----------------+ | 15| 284,700| 1. 19. 10. 15. 0| +--------+---------+-----------------+ | 16| 303,680| 2. 2. 3. 10. 0| +--------+---------+-----------------+ | 17| 322,660| 2. 4. 16. 5. 0| +--------+---------+-----------------+ | 18| 341,640| 2. 7. 9. 0. 0| +--------+---------+-----------------+ | 19| 360,620| 2. 10. 1. 13. 0| +--------+---------+-----------------+ | 20| 379,600| 2. 12. 14. 8. 0| +--------+---------+-----------------+ | 21| 398,580| 2. 15. 7. 3. 0| +--------+---------+-----------------+ | 22| 417,560| 2. 17. 19. 16. 0| +--------+---------+-----------------+ | 23| 436,540| 3. 0. 12. 11. 0| +--------+---------+-----------------+ | 24| 455,520| 3. 3. 5. 6. 0| +--------+---------+-----------------+ | 25| 474,500| 3. 5. 18. 1. 0| +--------+---------+-----------------+ | 26| 493,480| 3. 8. 10. 14. 0| +--------+---------+-----------------+ | 27| 512,460| 3. 11. 3. 9. 0| +--------+---------+-----------------+ | 28| 531,440| 3. 13. 16. 4. 0| +--------+---------+-----------------+ | 29| 550,420| 3. 16. 8. 17. 0| +--------+---------+-----------------+ | 30| 569,400| 3. 19. 1. 12. 0| +--------+---------+-----------------+ | 31| 588,380| 4. 1. 14. 7. 0| +--------+---------+-----------------+ | 32| 607,360| 4. 4. 7. 2. 0| +--------+---------+-----------------+ | 33| 626,340| 4. 6. 19. 15. 0| +--------+---------+-----------------+ | 34| 645,320| 4. 9. 12. 10. 0| +--------+---------+-----------------+ | 35| 664,300| 4. 12. 5. 5. 0| +--------+---------+-----------------+ | 36| 683,280| 4. 14. 18. 0. 0| +--------+---------+-----------------+ | 37| 702,260| 4. 17. 10. 13. 0| +--------+---------+-----------------+ | 38| 721,240| 5. 0. 3. 8. 0| +--------+---------+-----------------+ | 39| 740,220| 5. 2. 16. 3. 0| +--------+---------+-----------------+ | 40| 759,200| 5. 5. 8. 16. 0| +--------+---------+-----------------+ | 41| 778,180| 5. 8. 1. 11. 0| +--------+---------+-----------------+ | 42| 797,160| 5. 10. 14. 6. 0| +--------+---------+-----------------+ | 43| 816,140| 5. 13. 7. 1. 0| +--------+---------+-----------------+ | 44| 835,120| 5. 15. 19. 14. 0| +--------+---------+-----------------+ | 45| 854,100| 5. 18. 12. 9. 0| +--------+---------+-----------------+ | 46| 873,080| 6. 1. 5. 4. 0| +--------+---------+-----------------+ | 47| 892,060| 6. 3. 17. 17. 0| +--------+---------+-----------------+ | 48| 911,040| 6. 6. 10. 12. 0| +--------+---------+-----------------+ | 49| 930,020| 6. 9. 3. 7. 0| +--------+---------+-----------------+ | 50| 949,000| 6. 11. 16. 2. 0| +--------+---------+-----------------+ | 51| 967,980| 6. 14. 8. 15. 0| +--------+---------+-----------------+ | 52| 986,960| 6. 17. 1. 10. 0| +--------+---------+-----------------+ | 53|1,005,940| 6. 19. 14. 5. 0| +--------+---------+-----------------+ | 54|1,024,920| 7. 2. 7. 0. 0| +--------+---------+-----------------+ | 55|1,043,900| 7. 4. 19. 13. 0| +--------+---------+-----------------+ | 56|1,062,880| 7. 7. 12. 8. 0| +--------+---------+-----------------+ | 57|1,081,860| 7. 10. 5. 3. 0| +--------+---------+-----------------+ | 58|1,100,840| 7. 12. 17. 16. 0| +--------+---------+-----------------+ | 59|1,119,820| 7. 15. 10. 11. 0| +--------+---------+-----------------+ | 60|1,138,800| 7. 18. 3. 6. 0| +--------+---------+-----------------+ | 61|1,157,780| 8. 0. 16. 1. 0| +--------+---------+-----------------+ | 62|1,176,760| 8. 3. 8. 14. 0| +--------+---------+-----------------+ | 63|1,195,740| 8. 6. 1. 9. 0| +--------+---------+-----------------+ | 64|1,214,720| 8. 8. 14. 4. 0| +--------+---------+-----------------+ | 65|1,233,700| 8. 11. 6. 17. 0| +--------+---------+-----------------+ | 66|1,252,680| 8. 13. 19. 12. 0| +--------+---------+-----------------+ | 67|1,271,660| 8. 16. 12. 7. 0| +--------+---------+-----------------+ | 68|1,290,640| 8. 19. 5. 2. 0| +--------+---------+-----------------+ | 69|1,309,620| 9. 1. 17. 15. 0| +--------+---------+-----------------+ | 70|1,328,600| 9. 4. 10. 10. 0| +--------+---------+-----------------+ | 71|1,347,580| 9. 7. 3. 5. 0| +--------+---------+-----------------+ | 72|1,366,560| 9. 9. 16. 0. 0| +--------+---------+-----------------+ | 73|1,385,540| 9. 12. 8. 13. 0| +--------+---------+-----------------+ | 74|1,404,520| 9. 15. 1. 8. 0| +--------+---------+-----------------+ | 75|1,423,500| 9. 17. 14. 3. 0| +--------+---------+-----------------+ | 76|1,442,480|10. 0. 6. 16. 0| +--------+---------+-----------------+ | 77|1,461,460|10. 2. 19. 11. 0| +--------+---------+-----------------+ | 78|1,480,440|10. 5. 12. 6. 0| +--------+---------+-----------------+ | 79|1,499,420|10. 8. 5. 1. 0| +--------+---------+-----------------+ | 80|1,518,400|10. 10. 17. 14. 0| +--------+---------+-----------------+

{145}

Let us count forward the number 5,799 from the starting point 2 Kan 7 Tzec. It is apparent at the outset that, since this number is less than 18,980, or 1 Calendar Round, the preliminary rule given on page 143 does not apply in this case. Therefore we may proceed with the first rule given on page 139, by means of which the new day coefficient may be determined. Dividing the given number by 13 we have: 5,799 ÷ 13 = 446-1/13. Counting forward the numerator of the fractional part of the resulting quotient (1) from the day coefficient of the starting point (2), we reach 3 as the day coefficient of the terminal date.

The second rule given on page 140 tells how to find the day sign of the terminal date. Dividing the given number by 20, we have: 5,799 ÷ 20 = 289-19/20. Counting forward the numerator of the fractional part of the resulting quotient (19) from the day sign of the starting point, Kan, in the sequence of the twenty-day signs given in Table I, the day sign Akbal will be reached, which will be the day sign of the terminal date. Therefore the day of the terminal date will be 3 Akbal.

The third rule, given on page 141, tells how to find the position which the day of the terminal date occupied in the 365-day year. Dividing the given number by 365, we have: 5,799 ÷ 365 = 15-324/365. Counting forward the numerator of the fractional part of the resulting quotient, 324, from the year position of the starting date, 7 Tzec, in the sequence of the 365 year positions given in Table XV, the position 6 Zip will be reached as the position in the year of the day of the terminal date. The count by means of which the position 6 Zip is determined is given in detail. After the year position of the starting point, 7 Tzec, it requires 12 more positions (Nos. 8-19, inclusive) before the close of that month (see Table XV) will be reached. And after the close of Tzec, 13 uinals and the xma kaba kin must pass before the end of the year; 13 × 20 + 5 = 265, and 265 + 12 = 277. This latter number subtracted from 324, the total number of positions to be counted forward, will give the number of positions which remain to be counted in the next year following: 324 - 277 = 47. Counting forward 47 in the new year, we find that it will use up the months Pop and Uo (20 + 20 = 40) and extend 7 positions into the month Zip, or to 6 Zip. Therefore, gathering together the values determined for the several parts of the terminal date, we may say that in counting forward 5,799 from the starting point 2 Kan 7 Tzec, the terminal date reached will be 3 Akbal 6 Zip.

For the next example let us select a much higher number, say 322,920, which we will assume is to be counted forward from the starting point 13 Ik 0 Zip. Since this number is above 18,980, we may apply our preliminary rule (p. 143) and deduct all the Calendar {146} 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 0 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 we have 260 ÷ 20 = 13. Since there is no fraction in the quotient, the numerator of the fraction will be 0, and counting forward 0 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.[101] Dividing by 365 we have: 260 ÷ 365 = 260/365. Counting forward the numerator of the fraction, 260, from the year position of the starting point, 0 Zip, in Table XV, the position in the year of the day of the terminal date will be found to be 0 Pax. Since 260 days equal just 13 uinals, we have only to count forward from 0 Zip 13 uinals in order to reach the year position; that is, 0 Zotz is 1 uinal; to 0 Tzec 2 uinals, to 0 Xul 3 uinals, and so on in Table XV to 0 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 0 Zip, the terminal date reached is 13 Ik 0 Pax.

In order to illustrate the method of procedure when the count is _backward_, let us assume an example of this kind. 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 number by 13, we have: 9,663 ÷ 13 =743-4/13. Counting the numerator of the fractional part of this quotient, 4, _backward_ from the day coefficient of the starting point, 3, we reach 12 as the day coefficient of the terminal date, that is, 2, 1, 13, 12 (rule 1, p. 139). Dividing the given number by 20, we have: 9,663 ÷ 20 = 483-3/20. 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 will be 12 Eznab. Dividing the given number by 365, we have {147} 9,663 ÷ 365 = 26-173/365. Counting _backward_ the numerator of the fractional part of this quotient, 173, from the year position of the starting point, 4 Uayeb, the year position of the terminal date will be found to be 11 Yax. Before position 4 Uayeb (see Table XV) there are 4 positions in that division of the year (3, 2, 1, 0). Counting these _backward_ to the end of the month Cumhu (see Table XV), we have left 169 positions (173 - 4 = 169); this equals 8 uinals and 9 days extra. Therefore, beginning with the end of Cumhu, we may count _backward_ 8 whole uinals, namely: Cumhu, Kayab, Pax, Muan, Kankin, Mac, Ceh, and Zac, which will bring us to the end of Yax (since we are counting backward). As we have left still 9 days out of our original 173, these must be counted backward from position 0 Zac, that is, beginning with position 19 Yax: 19, 18, 17, 16, 15, 14, 13, 12, 11; so 11 Yax is the position in the year of the day of the terminal date. Assembling the above values, we find that in counting the number 9,663 _backward_ from the starting point, 2 Imix 4 Uayeb, the terminal date is 12 Eznab 11 Yax. Whether the count be forward or backward, the method is the same, the only difference being in the direction of the counting.

This concludes the discussion of the actual arithmetical processes involved in counting forward or backward any given number from any given date; however, before explaining the fifth and final step in deciphering the Maya numbers, it is first necessary to show how this method of counting was applied to the Long Count.

The numbers used above in connection with dates merely express the difference in time between starting points and terminal dates, without assigning either set of dates to their proper positions in Maya chronology; that is, in the Long Count. Consequently, since any Maya date recurred at successive intervals of 52 years, by the time their historic period had been reached, more than 3,000 years after the starting point of their chronology, the Maya had upward of 70 distinct dates of exactly the same name to distinguish from one another.

It was stated on page 61 that the 0, or starting point of Maya chronology, was the date 4 Ahau 8 Cumhu, from which all subsequent dates were reckoned; and further, on page 63, that by recording the number of cycles, katuns, tuns, uinals, and kins which had elapsed in each case between this date and any subsequent dates in the Long Count, subsequent dates of the same name could be readily distinguished from one another and assigned at the same time to their proper positions in Maya chronology. This method of fixing a date in the Long Count has been designated Initial-series dating.

The generally accepted method of writing Initial Series is as follows:

9.0.0.0.0. 8 Ahau 13 Ceh

The particular Initial-Series written here is to be interpreted thus: "Counting forward 9 cycles, 0 katuns, 0 tuns, 0 uinals, and 0 kins {148} from 4 Ahau 8 Cumhu, the starting point of Maya chronology (always unexpressed in Initial Series), the terminal date reached will be 8 Ahau 13 Ceh."[102] Or again:

9.14.13.4.17. 12 Caban 5 Kayab

This Inital Series reads thus: "Counting forward 9 cycles, 14 katuns, 13 tuns, 4 uinals, and 17 kins from 4 Ahau 8 Cumhu, the starting point of Maya chronology (unexpressed), the terminal date reached will be 12 Caban 5 Kayab."

The time which separates any date from 4 Ahau 8 Cumhu may be called that date's Initial-series value. For example, in the first of the above cases the number 9.0.0.0.0 is the Initial-series value of the date 8 Ahau 13 Ceh, and in the second the number 9.14.13.4.17 is the Initial-series value of the date 12 Caban 5 Kayab. It is clear from the foregoing that although the date 8 Ahau 13 Ceh, for example, had recurred upward of 70 times since the beginning of their chronology, the Maya were able to distinguish any

## particular 8 Ahau 13 Ceh from all the others merely by recording its

distance from the starting point; in other words, giving thereto its

## particular Initial-series value, as 9.0.0.0.0. in the present case.

Similarly, any particular 12 Caban 5 Kayab, by the addition of its corresponding Initial-series value, as 9.14.13.4.17 in the case above cited, was absolutely fixed in the Long Count--that is, in a period of 374,400 years.

Returning now to the question of how the counting of numbers was applied to the Long Count, it is evident that _every date in Maya chronology, starting points as well as terminal dates, had its own particular Initial-series value_, though in many cases these values are not recorded. However, in most of the cases in which the Initial-series values of dates are not recorded, they may be calculated by means of their distances from other dates, whose Initial-series values are known. This adding and subtracting of numbers to and from Initial Series[103] constitutes the application of the above-described arithmetical processes to the Long Count. Several examples of this use are given below.

Let us assume for the first case that the number 2.5.6.1 is to be counted forward from the Initial Series 9.0.0.0.0 8 Ahau 13 Ceh. By multiplying the values of the katuns, tuns, uinals, and kins given in Table XIII by their corresponding coefficients, in this case 2, 5, 6, and 1, respectively, and adding the resulting products together, we find that 2.5.6.1 reduces to 16,321 units of the first order.

Counting this forward from 8 Ahau 13 Ceh as indicated by the rules on pages 138-143, the terminal date 1 Imix 9 Yaxkin will be reached. {149} Moreover, since the Initial-series value of the starting point 8 Ahau 13 Ceh was 9.0.0.0.0, the Initial-series value of 1 Imix 9 Yaxkin, the terminal date, may be calculated by adding its distance from 8 Ahau 13 Ceh to the Initial-series value of that date:

9.0.0.0.0 (Initial-series value of starting point) 8 Ahau 13 Ceh 2.5.6.1 (distance from 8 Ahau 13 Ceh to 1 Imix 9 Yaxkin) 9.2.5.6.1 (Initial-series value of terminal date) 1 Imix 9 Yaxkin

That is, by calculation we have determined the Initial-series value of the

## particular 1 Imix 9 Yaxkin, which was distant 2.5.6.1 from 9.0.0.0.0 8 Ahau

13 Ceh, to be 9.2.5.6.1, notwithstanding that this fact was not recorded.

The student may prove the accuracy of this calculation by treating 9.2.5.6.1 1 Imix 9 Yaxkin as a new Initial Series and counting forward 9.2.5.6.1 from 4 Ahau 8 Cumhu, the starting point of all Initial Series known except two. If our calculations are correct, the former date will be reached just as if we had counted forward only 2.5.6.1 from 9.0.0.0.0 8 Ahau 13 Ceh.

In the above example the distance number 2.5.6.1 and the date 1 Imix 9 Yaxkin to which it reaches, together are called a Secondary Series. This method of dating already described (see pp. 74-76 et seq.) seems to have been used to avoid the repetition of the Initial-series values for all the dates in an inscription. For example, in the accompanying text--

9.12. 2. 0.16 5 Cib 14 Yaxkin 12. 9.15 [9.12.14.10.11][104] 9 Chuen 9 Kankin 5 [9.12.14.10.16] 1 Cib 14 Kankin 1. 0. 2. 5 [9.13.14.13. 1] 5 Imix 19 Zac

the only parts actually recorded are the Initial Series 9.12.2.0.16 {150} 5 Cib 14 Yaxkin, and the Secondary Series 12.9.15 leading to 9 Chuen 9 Kankin; the Secondary Series 5 leading to 1 Cib 14 Kankin; and the Secondary Series 1.0.2.5 leading to 5 Imix 19 Zac. The Initial-series values: 9.12.14.10.11; 9.12.14.10.16; and 9.13.14.13.1, belonging to the three dates of the Secondary Series, respectively, do not appear in the text at all (a fact indicated by the brackets), but are found only by calculation. Moreover, the student should note that in a succession of interdependent series like the ones just given the terminal date reached by one number, as 9 Chuen 9 Kankin, becomes the starting point for the next number, 5. Again, the terminal date reached by counting 5 from 9 Chuen 9 Kankin, that is, 1 Cib 14 Kankin, becomes the starting point from which the next number, 1.0.2.5, is counted. In other words, these terms are only relative, since the terminal date of one number will be the starting point of the next.

Let us assume for the next example, that the number 3.2 is to be counted forward from the Initial Series 9.12.3.14.0 5 Ahau 8 Uo. Reducing 3 uinals and 2 kins to kins, we have 62 units of the first order. Counting forward 62 from 5 Ahau 8 Uo, as indicated by the rules on pages 138-143, it is found that the terminal date will be 2 Ik 10 Tzec. Since the Initial-series value of the starting point 5 Ahau 8 Uo is known, namely, 9.12.3.14.0, the Initial Series corresponding to the terminal date may be calculated from it as before:

9.12.3.14.0 (Initial-series value of the starting point) 5 Ahau 8 Uo 3.2 (distance from 5 Ahau 8 Uo forward to 2 Ik 10 Tzec) [9.12.3.17.2] (Initial-series value of the terminal date) 2 Ik 10 Tzec

The bracketed 9.12.3.17.2 in the Initial-series value corresponding to the date 2 Ik 10 Tzec does not appear in the record but was reached by calculation. The student may prove the accuracy of this result by treating 9.12.3.17.2 2 Ik 10 Tzec as a new Initial Series, and counting forward 9.12.3.17.2 from 4 Ahau 8 Cumhu (the starting point of Maya chronology, unexpressed in Initial Series). If our calculations are correct, the same date, 2 Ik 10 Tzec, will be reached, as though we had counted only 3.2 forward from the Initial Series 9.12.3.14.0 5 Ahau 8 Uo.

One more example presenting a "backward count" will suffice to illustrate this method. Let us count the number 14.13.4.17 _backward_ from the Initial Series 9.14.13.4.17 12 Caban 5 Kayab. Reducing 14.13.4.17 to units of the 1st order, we have 105,577. Counting this number _backward_ from 12 Caban 5 Kayab, as indicated in the rules on pages 138-143, we find that the terminal date will be 8 Ahau 13 Ceh. Moreover, since the Initial-series value of the starting point 12 Caban 5 Kayab is known, namely, 9.14.13.4.17, the Initial-series value of {151} the terminal date may be calculated by _subtracting_ the distance number 14.13.4.17 from the Initial Series of the starting point:

9.14.13.4.17 (Initial-series value of the starting point) 12 Caban 5 Kayab 14.13.4.17 (distance from 12 Caban 5 Kayab backward to 8 Ahau 13 Ceh) [9. 0. 0.0. 0] (Initial-series value of the terminal date) 8 Ahau 13 Ceh

The bracketed parts are not expressed. We have seen elsewhere that the Initial Series 9.0.0.0.0 has for its terminal date 8 Ahau 13 Ceh; therefore our calculation proves itself.

The foregoing examples make it sufficiently clear that the distance numbers of Secondary Series may be used to determine the Initial-series values of Secondary-series dates, either by their addition to or subtraction from known Initial-series dates.

We have come now to the final step in the consideration of Maya numbers, namely, the identification of the terminal dates determined by the calculations given under the fourth step, pages 138-143. This step may be summed up as follows:

FIFTH STEP IN SOLVING MAYA NUMBERS

Find the terminal date to which the number leads.

As explained under the fourth step (pp. 138-143), the terminal date may be found by calculation. The above direction, however, refers to the actual finding of the terminal dates in the texts; that is, where to look for them. It may be said at the outset in this connection that terminal dates in the great majority of cases follow immediately the numbers which lead to them. Indeed, the connection between distance numbers and their corresponding terminal dates is far closer than between distance numbers and their corresponding starting points. This probably results from the fact that the closing dates of Maya periods were of far more importance than their opening dates. Time was measured by elapsed periods and recorded in terms of the ending days of such periods. The great emphasis on the closing date of a period in comparison with its opening date probably caused the suppression and omission of the date 4 Ahau 8 Cumhu, the starting point of Maya chronology, in all Initial Series. To the same cause also may probably be attributed the great uniformity in the positions of almost all terminal dates, i.e., immediately after the numbers leading to them.

We may formulate, therefore, the following general rule, which the student will do well to apply in every case, since exceptions to it are very rare:

_Rule._ The terminal date reached by a number or series almost invariably follows immediately the last term of the number or series leading to it. {152}

This applies equally to all terminal dates, whether in Initial Series, Secondary Series, Calendar-round dating or Period-ending dating, though in the case of Initial Series a peculiar division or partition of the terminal date is to be noted.

Throughout the inscriptions, excepting in the case of Initial Series, the month parts of the dates almost invariably follow immediately the days whose positions in the year they designate, without any other glyphs standing between; as, for example, 8 Ahau 13 Ceh, 12 Caban 5 Kayab, etc. In Initial Series, on the other hand, the day parts of the dates, as 8 Ahau and 12 Caban, in the above examples, are almost invariably separated from their corresponding month parts, 13 Ceh or 5 Kayab, by several intervening glyphs. The positions of the day parts in Initial-series terminal dates are quite regular according to the terms of the above rule; that is, they follow immediately the lowest period of the number which in each case shows their distance from the unexpressed starting point, 4 Ahau 8 Cumhu. The positions of the corresponding month parts are, on the other hand, irregular. These, instead of standing immediately after the days whose positions in the year they designate, follow at the close of some six or seven intervening glyphs. These intervening glyphs have been called the Supplementary Series, though the count which they record has not as yet been deciphered.[105] The month glyph in the great majority of cases follows immediately the closing[106] glyph of the Supplementary Series. The form of this latter sign is always unmistakable (see fig. 65), and it is further characterized by its numerical coefficient, which can never be anything but 9 or 10.[107] See examples of this sign in the figure just mentioned, where both normal forms _a, c, e, g,_ and _h_ and head variants _b, d,_ and _f_ are included.

The student will find this glyph exceedingly helpful in locating the month parts of Initial-series terminal dates in the inscriptions. For example, let us suppose in deciphering the Initial Series 9.16.5.0.0 8 Ahau 8 Zotz that the number 9.16.5.0.0 has been counted forward {153} from 4 Ahau 8 Cumhu (the unexpressed starting point), and has been found by calculation to reach the terminal date 8 Ahau 8 Zotz; and further, let us suppose that on inspecting the text the day part of this date (8 Ahau) has been found to be recorded immediately after the 0 kins of the number 9.16.5.0.0. Now, if the student will follow the next six or seven glyphs until he finds one like any of the forms in figure 65, the glyph immediately following the latter sign will be in all probability the month part, 8 Zotz in the above example, of an Initial-series' terminal date. In other words, although the meaning of the glyph shown in the last-mentioned figure is unknown, it is important for the student to recognize its form, since it is almost invariably the "indicator" of the month sign in Initial Series.

[Illustration: FIG. 65. Sign for the "month indicator": _a, c, e, g, h_, Normal forms; _b, d, f_, head variants.]

In all other cases in the inscriptions, including also the exceptions to the above rule, that is, where the month parts of Initial-series terminal dates do not immediately follow the closing glyph of the Supplementary Series, the month signs follow immediately the day signs whose positions in the year they severally designate.

In the codices the month signs when recorded[108] usually follow immediately the days signs to which they belong. The most notable exception[109] to this general rule occurs in connection with the Venus-solar periods represented on pages 46-50 of the Dresden Codex, where one set of day signs is used with three different sets of month signs to form three different sets of dates. For example, in one place the day 2 Ahau stands above three different month signs--3 Cumhu, 3 Zotz, and 13 Yax--with each of which it is used to form a {154} different date--2 Ahau 3 Cumhu, 2 Ahau 3 Zotz, and 2 Ahau 13 Yax. In these pages the month signs, with a few exceptions, do not follow immediately the days to which they belong, but on the contrary they are separated from them by several intervening glyphs. This abbreviation in the record of these dates was doubtless prompted by the desire or necessity for economizing space. In the above example, instead of repeating the 2 Ahau with each of the two lower month signs, 3 Zotz and 13 Yax, by writing it once above the upper month sign, 3 Cumhu, the scribe intended that it should be used in turn with each one of the three month signs standing below it, to form three different dates, saving by this abbreviation the space of two glyphs, that is, double the space occupied by 2 Ahau.

With the exception of the Initial-series dates in the inscriptions and the Venus-Solar dates on pages 46-50 of the Dresden Codex, we may say that the regular position of the month glyphs in Maya writing was immediately following the day glyphs whose positions in the year they severally designated.

In closing the presentation of this last step in the process of deciphering numbers in the texts, the great value of the terminal date as a final check for all the calculations involved under steps 1-4 (pp. 134-151) should be pointed out. If after having worked out the terminal date of a given number according to these rules the terminal date thus found should differ from that actually recorded under step 5, we must accept one of the following alternatives:

1. There is an error in our own calculations; or

2. There is an error in the original text; or

3. The case in point lies without the operation of our rules.

It is always safe for the beginner to proceed on the assumption that the first of the above alternatives is the cause of the error; in other words, that his own calculations are at fault. If the terminal date as calculated does not agree with the terminal-date as recorded, the student should repeat his calculations several times, checking up each operation in order to eliminate the possibility of a purely arithmetical error, as a mistake in multiplication. After all attempts to reach the recorded terminal date by counting the given number from the starting point have failed, the process should be reversed and the attempt made to reach the starting point by counting backward the given number from its recorded terminal date. Sometimes this reverse process will work out correctly, showing that there must be some arithmetical error in our original calculations which we have failed to detect. However, when both processes have failed several times to connect the starting point with the recorded terminal date by use of the given number, there remains the possibility that either the starting point or the terminal date, or perhaps both, do not belong to the given number. The rules for determining this fact {155} have been given under step 2, page 135, and step 4, page 138. If after applying these to the case in point it seems certain that the starting point and terminal date used in the calculations both belong to the given number, we have to fall back on the second of the above alternatives, that is, that there is an error in the original text.

Although very unusual, particularly in the inscriptions, errors in the original texts are by no means entirely unknown. These seem to be restricted chiefly to errors in numerals, as the record of 7 for 8, or 7 for 12 or 17, that is, the omission or insertion of one or more bars or dots. In a very few instances there seem to be errors in the month glyph. Such errors usually are obvious, as will be pointed out in connection with the texts in which they are found (see Chapters V and VI).

If both of the above alternatives are found not to apply, that is, if both our calculations and the original texts are free from error, we are obliged to accept the third alternative as the source of trouble, namely, that the case in point lies without the operation of our rules. In such cases it is obviously impossible to go further in the present state of our knowledge. Special conditions presented by glyphs whose meanings are unknown may govern such cases. At all events, the failure of the rules under 1-4 to reach the terminal dates recorded as under 5 introduces a new phase of glyph study--the meaning of unknown forms with which the beginner has no concern. Consequently, when a text falls without the operation of the rules given in this chapter--a very rare contingency--the beginner should turn his attention elsewhere. {156}