126 



MEMOIR EXPLANATORY OF A NEW PERPETUAL CALENDAR. 



3d Rule. As a means of proving all the foregoing operations, or in lieu of them, with a view, 

 chiefly, to save figures in computing the Epact, employ the following formulas, which do 

 not refer either to the cycle of the sun, or to that of the Golden Numbers, now in common use, 

 beo-innino- with the year 1 + 9, and the year 1 -(- 1, respectively, but to a succession of cycles 

 of 28 and 19, in which the years corresponding with the first year of the Christian Era 

 must always be numbered the jirst of each cycle also. 



FORMULAE. 



Call any astronomical year before Christ, y > ^ theif Reraain(]er after divigion b either , 

 and any A. D., or year after Christ, l } j j 



If r = 0, keep it so; except in Old Style Years, when, if 19 be the Divisor, change to 19. 



Then 28 — (^) i an( 3 19 — ("To) ■ W "H express s/'s No. in each cycle in the Era B. C. 



And (— ) , and (lo) . > W 'H express F's No. in each cycle in the Era after Christ. 



Apply the Rules of the Tablet, on each side, to the Cyclic No. thus found, as if it were the given year, using 

 always the secular equations A and C, belonging to the given Era, and the results will be uniformly the same, 

 as in the examples heretofore stated. 



When the Cyclic No. on the Civil side is a multiple of 4, the year is leap, unless it be a New Style 100th year 

 unmarked with an asterisk. 



When the Cyclic No. on the Church side is 19, " 1 less than the 19th part" becomes 0, and the Exception to 

 the Rule for finding the Epact is thus eliminated. 



EXAMPLES. 

 1. Required both the Old and New Style Easter of A. 1). 1848, (being a multiple of 28.) 



Julian Year. 



28) 1848 

 Remainder 

 its 4th 



A 5 



Month 

 Day 



6 

 10 



21Rem. 



Saturday, or 7 

 From 8 



Easter. 



19) 1848 

 Remainder 5 

 5 X 10 = 50 

 C 



30) 55 



Epact 25 



From 35 



JTCrtn April 10 



. to Sunday 1 

 Answer, April 11 



Gregorian Year* 



28) 1848 

 Remainder 

 its 4 th 



A 



Month 

 Day 







18 



7) 24 Rem. 



Tuesday 

 From 



Easter. 



19) 1848 

 Remainder 5 

 5 X 10 = 50 

 C 



30) 55 



Epact 25 



From 43 



♦Term April is 



5 to Sunday 5 



Answer, April 23 



2. Required both the Old and New Style Easter of A. D. 2698, (being a multiple of 19.) 



Julian Year. 



28) 2G98 

 Remainder 10 

 its 4th 2 



A 5 



Month 



Day 5 



7) 28 Rem. 



Saturday or 7 



From 8 



Easter. 



19) 2698 

 Remainder = 

 Divisor 19 

 19x10=190 

 C 



30) 209 



Epact 

 From 



JTcrm April 



. . to Sunday, 



Answer, April 



29 

 34 



Gregorian Year. 



28) 2698 

 Remainder 10 

 its 4th 2 



A 1 



Month 

 Day 



6 

 17 



7) 36 Rem. 

 Sunday 1 



From 8 



7 



Easter. 



19) 2698 

 Remainder 0(kept0) 

 0X10=0 

 C 26 



30) 26 



Epact 26 

 From 43 



armn April 17 



. to Sunday 7 

 Answer, April 24 



