Page:A budget of paradoxes (IA cu31924103990507).pdf/238

224 festival after the Jews, &c., more than astronomical correctness. He gives comparison tables which would startle a believer in the astronomical intention of his Calendar: they are to show that a calendar in which the moon is always made a day older than by him, represents the heavens better than he has done, or meant to do. But it must be observed that this diminution of the real moon's age has a tendency to make the English explanation often practically accordant with the Calendar. For the fourteenth day of Clavius is generally the fifteenth day of the mean moon of the heavens, and therefore most often that of the real moon. But for this, 1818 and 1845 would not have been the only instances of our day in which the English precept would have contradicted the Calendar.

11. In the construction of the Calendar, Clavius adopted the ancient cycle of 532 years, but, we may say, without ever allowing it to run out. At certain periods, a shift is made from one part of the cycle into another. This is done whenever what should be Julian leap year is made a common year, as in 1700, 1800, 1900, 2100, &c. It is also done at certain times to correct the error of 1h. 19m., before referred to, in each cycle of golden numbers: Clavius, to meet his view of the amount of that error, put forward the moon's age a day 8 times in 2,500 years. As we cannot enter at full length into the explanation, we must content ourselves with giving a set of rules, independent of tables, by which the reader may find Easter for himself in any year, either by the old Calendar or the new. Any one who has much occasion to find Hasters and moveable feasts should procure Franecœur's tables.

12. Rule for determining Easter Day of the Gregorian Calendar in any year of the new style. To the several parts of the rule are annexed, by way of example, the results for the year 1849.

I. Add 1 to the given year. (1850).

II. Take the quotient of the given year divided by 4, neglecting the remainder. (462).

III. Take 16 from the centurial figures of the given year, if it can be done, and take the remainder. (2).

IV. Take the quotient of III. divided by 4, neglecting the remainder. (0).

V. From the sum of I., II., and IV., substract III. (2310).

VI. Find the remainder of V. divided by 7. (0).

VII. Subtract VI. from 7; this is the number of the dominical letter

VIII. Divide I. by 19, the remainder (or 19, if no remainder) is the golden number. (7).