MORLEY] INTRODUCTION TO STUDY OF MAYA HIEEOGLYPHS 149 



Moreover, since the Initial-series value of the starting pomt 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 coimted 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. Tliis 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 accom- 

 panying text — 



9.12. 2. 0.16 5 Cib 14 Yaxkin 



12. 9.15 

 [9.12.14.10.11] ^ 9 Chuen 9 Kankin 



5 

 [9.12.14.10.16] 1 Cib 14 Kankin 



1. 0. 2. 5 

 [9.13.14.13. 1] 6 Imix 19 Zac 



1 In adding two Maya numbers, for example 9.12.2.0.16 and 12.9.5, care should be taken first to arrange 

 like units under like, as: 



9.12. 2. 0.16 

 12. 9. 5 



9.12.14.10. 1 

 Next, begraning at the right, the kins or units of the 1st place are added together, and after all the 20s 

 (here 1) have been deducted from this sum, place the remainder (here 1) in the kin place. Next add the 

 uinals, or units of the 2d place, adding to them 1 for each 20 which was carried forward from the 1st place. 

 After all the 18s possible have been deducted from this sum (here 0) place the remainder (here 10) in the 

 uinal place. Next add the tuns, or units of the 3d place, adding to them 1 for each 18 which was carried 

 forward from the 2d place, and after deducting all the 20s possible (here 0) place the remainder (here 14) 

 in the tun place. Proceed in this manner until the highest units present have been added and written 

 below. 

 Subtraction is just the reverse of the preceding. Using the same numbers: 



9.12. 2.0.16 

 12.9. 5 



9.11. 9.9.11 

 5 kins from 16=11; 9 uinals from 18 uinals (1 tunhas to be borrowed)=9; 12 tims from 21 tuns (1 katun has 

 to be borrowed, which, added to the 1 tun left in the minuend, makes 21 tuns)=9 tuns; katuns from 

 11 katuns (1 katun having been borrowed)= 11 katims; and cycles from 9 cycles= 9 cycles. 



