Fig. 62Fig.62. Glyphs showing misplacement of the kin coefficient (a) or elimination of a period glyph (b,c):a, Stela E, Quirigua;b, Altar U, Copan;c, Stela J, Copan.
Fig.62. Glyphs showing misplacement of the kin coefficient (a) or elimination of a period glyph (b,c):a, Stela E, Quirigua;b, Altar U, Copan;c, Stela J, Copan.
The omission of the kin sign, while by far the most common, is not the only example of glyph omission found in numerical series in the inscriptions. Sometimes, though very rarely, numbers occur in which periods other than the kin are wanting. A case in point is figure62,b. Here a tun sign appears with the coefficient 13 above and 3 to the left. Since there are only two coefficients (13 and 3) and three time periods (tun, uinal, and kin), it is clear that the signs of both the lower periods have been omitted as well as the coefficient of one of them. Incof the last-mentioned figure a somewhat different practice was followed. Here, although three time periods are recorded—tuns, uinals and kins—one period (the uinal) and its coefficient have been omitted, and there is nothing between the 0 kins and 10 tuns. Such cases are exceedingly rare, however, and may be disregarded by the beginner.
We have seen that the order of the periods in the numbers in figure56was just the reverse of that in the numbers shown in figures58and59; that in one place the kins stand at the top and in the other at the bottom; and finally, that this difference was not a vital one, since it had no effect on the values of the numbers. This is true, because in the first method of expressing the higher numbers, it matters not which end of the number comes first, the highest or thelowest period, so long as its several periods always stand in the same relation to each other. For example, in figure56,q, 6 cycles, 17 katuns, 2 tuns, 10 uinals, and 0 kins represent exactly the same number as 0 kins, 10 uinals, 2 tuns, 17 katuns, and 6 cycles; that is, with the lowest term first.
It was explained on page23that the order in which the glyphs are to be read is from top to bottom and from left to right. Applying this rule to the inscriptions, the student will find that all Initial Series are descending series; that in reading from top to bottom and left to right, the cycles will be encountered first, the katuns next, the tuns next, the uinals, and the kins last. Moreover, it will be found also that the great majority of Secondary Series are ascending series, that is, in reading from top to bottom and left to right, the kins will be encountered first, the uinals next, the tuns next, the katuns next, and the cycles last. The reason why Initial Series always should be presented as descending series, and Secondary Series usually as ascending series is unknown; though as stated above, the order in either case might have been reversed without affecting in any way the numerical value of either series.
This concludes the discussion of the first method of expressing the higher numbers, the only method which has been found in the inscriptions.
Second Method of Numeration
The other method by means of which the Maya expressed their higher numbers (the second method given on p.103) may be called "numeration by position," since in this method the numerical value of the symbols depended solely on position, just as in our own decimal system, in which the value of a figure depends on its distance from the decimal point, whole numbers being written to the left and fractions to the right. The ratio of increase, as the word "decimal" implies, is 10 throughout, and the numerical values of the consecutive positions increase as they recede from the decimal point in each direction, according to the terms of a geometrical progression. For example, in the number 8888.0, the second 8 from the decimal point, counting from right to left, has a value ten times greater than the first 8, since it stands for 8 tens (80); the third 8 from the decimal point similarly has a value ten times greater than the second 8, since it stands for 8 hundreds (800); finally, the fourth 8 has a value ten times greater than the third 8, since it stands for 8 thousands (8,000). Hence, although the figures used are the same in each case, each has a different numerical value, depending solely upon its position with reference to the decimal point.
In the second method of writing their numbers the Maya had devised a somewhat similar notation. Their ratio of increase was 20 in all positions except the third. The value of these positions increasedwith their distance from the bottom, according to the terms of the vigesimal system shown in TableVIII. This second method, or "numeration by position," as it may be called, was a distinct advance over the first, since it required for its expression only the signs for the numerals 0 to 19, inclusive, and did not involve the use of any period glyphs, as did the first method. To its greater brevity, no doubt, may be ascribed its use in the codices, where numerical calculations running into numbers of 5 and 6 terms form a large part of the subject matter. It should be remembered that in numeration by position only the normal forms of the numbers—bar and dot numerals—are used. This probably results from the fact that head-variant numerals never occur independently, but are always prefixed to some other glyph, as period, day, or month signs (see p.104). Since no period glyphs are used in numeration by position, only normal-form numerals, that is, bar and dot numerals, can appear.
The numbers from 1 to 19, inclusive, are expressed in this method, as shown in figure39, and the number 0 as shown in figure46. As all of these numbers are below 20, they are expressed as units of the first place or order, and consequently each should be regarded as having been multiplied by 1, the numerical value of the first or lowest position.
The number 20 was expressed in two different ways: (1) By the sign shown in figure45; and (2) by the numeral 0 in the bottom place and the numeral 1 in the next place above it, as in figure63,a. The first of these had only a very restricted use in connection with the tonalamatl, wherein numeration by position was impossible, and therefore a special character for 20 (see fig.45) was necessary. See 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 TableVIII, we saw had numerical values of 1 and 20, respectively. For example, the number 37 was expressed as shown in figure63,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 figure63,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 TableVIII.) For example, the number 360 is shown in figure63,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 figure63,e.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),
Fig. 63Fig.63. Examples of the second method of numeration, used exclusively in the codices.
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 TableVIII.) For example, the number 7,202 is recorded in figure63,f.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 figure63,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 TableVIII.) For example, the number 169,200 is recorded in figure63,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 figure63,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 figure63,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 figure63,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).
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:
TableXII. COMPARISON OF THE TWO METHODS OF NUMERATION
We have seen in the foregoing pages (1) how the Maya wrote their 20numerals, 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 thefirst stepin 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 pages105-133, 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 TableXIIIis to be used for reducing numbers to their primary units in the inscriptions, and TableXIVfor the same purpose in the codices.
TableXIII. VALUES OF HIGHER PERIODS IN TERMS OF LOWEST, IN INSCRIPTIONS
TableXIV. VALUES OF HIGHER PERIODS IN TERMS OF LOWEST, IN CODICES
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 TableXIII). 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 TableXIV, 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 thesecond 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 standbetween.[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 be4 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 date4 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 isforwardfrom the starting point to somelater date, or whether it isbackwardfrom the starting point to someearlier date. This process may be called thethird 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 countedforwardfrom their starting points and not backward. In other words, they proceed fromearlier to later datesand 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.
Fig. 64Fig.64. Figure showing the use of the "minus" or "backward" sign in the codices.
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 figure64. 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 figure64the 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 figure64is counted will be found to be below it, that is, the date4 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 countedbackwardfrom 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 page136, 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 isalways forward, and, in general throughout the inscriptions. The very few cases in which the countisbackward, are confined chiefly to Secondary Series, and it is indealing 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, as4 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 TableI.
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 TableIII.
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.
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 thewhole-number partof the resulting quotient by 13: 31,741 ÷ 13 = 2,4418⁄13. 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 thenumeratorof the fractional part.[100]Hence the following rule for determining the first unknown on page138(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 quotient8⁄13. 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 date4 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 page37that the twenty day signs given in TableIsucceed 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.
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 = 1,5871⁄20. 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 TableI, 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 page138(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 TableI, 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 TableIfrom 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 fromAhauin TableI, the day signImixis reached, andImix, 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 andImixfor the second, we can now say that in counting forward 31,741 from the date4 Ahau 8 Cumhu, the day reached will be12 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 page138. Both of these may be found at one time by the same operation.
It was explained on pages44-51that 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 thexma kaba kinof 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 whencounted 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 = 86351⁄365. 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 TableXV, 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 page138(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 TableXV, 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.
TableXV. THE 365 POSITIONS IN THE MAYA YEAR
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 TableXVfrom the position8 Cumhu, the position of the day of the starting point,4 Ahau 8 Cumhu.
A glance at the month ofCumhuin TableXVshows that after the position8 Cumhuthere are 11 positions in that month; adding to these the 5 inUayeb, 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 TableXVfrom 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 TableXVhe will see that 15 positions counted forward in this month will reach the position14 Kayab, which is also the position reached by counting forward 31,741 positions from the starting position8 Cumhu.
Having determined values for all of the unknowns on page138, we can now say that if the number 31,741 be counted forward from the date4 Ahau 8 Cumhu, the date12 Imix 14 Kayabwill 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 isbackward, 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 pages51-60that 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.
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 TableXVIare 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 TableXVIthat 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 point4 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 = 9818⁄1312,761 ÷ 20 = 6381⁄2012,761 ÷ 365 = 34351⁄365
12,761 ÷ 13 = 9818⁄1312,761 ÷ 20 = 6381⁄2012,761 ÷ 365 = 34351⁄365
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 TableXVI), from the first, and 3 Calendar Rounds, 56,940 (see TableXVI), 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.
TableXVI. 80 CALENDAR ROUNDS EXPRESSED IN ARABIC AND MAYA NOTATION