By the Julian rule three-fourths of a day is gained every century, which in 400 years amounts to three days. This is corrected by the Gregorian, by making three consecutive centurial years common years, thus suppressing three days in 400 years. RULE. Multiply the difference between the Julian and the solar year by 100, and we have the error in 100 years. Multiply this product by 4 and we have the error in 400 years. Now, 400 is the tenth of 4,000; therefore, multiply the last product by 10, and we have the error in 4,000 years. Now, as the discrepancy between the Julian and Gregorian year is three days in 400 years, making 3-400 of a day every year, so by dividing 365-1/4, the number of days in a year, by 3-400, we have the time it would take to make a revolution of the seasons. SOLUTION. (365 d, 6 h.) - (365 d, 5 h, 48 m, 49.62 s.) = (11 m, 10.38 s.) Now, (11 m, 10.38 s.) x 100 = 18 h, 37.3 m, the gain in 100 years. This is, reckoned in round numbers, 18 hours, or three-fourths of a day. Now, (3/4 x 4) = (1 x 3) = 3: the Julian rule gaining three days, the Gregorian suppressing three days in 400 years. (3 x 10) = 30, the number of days gained by the Julian rule in 4,000 years. 365-1/4 / 3 400 = 48,700, so that in this long period of time, this falling back 3/4 of a day every century would amount to 365-1/4 days; therefore, 48,699 Julian years are equal to 48,700 Gregorian years. CHAPTER II. ERRORS OF THE GREGORIAN CALENDAR. By reference to the preceding chapter it will be seen that there is an error of 37.3 minutes in every 100 years not corrected by the Gregorian calendar; this amounts to only .373 of a minute a year, or one day in 3,861 years, and one day and fifty-two minutes in 4,000 years. RULE. To find how long it would take to gain one day: Divide the number of minutes in a day by the decimal .373, that being the fraction of a minute gained every year. To find how much time would be gained in 4,000 years, multiply the decimal .373 by 4,000, and you will have the answer in minutes, which must be reduced to hours. SOLUTION. (24 x 60) / .373 = 3,861, nearly; hence the error would amount to only one day in 3,861 years. (.373 x 4,000) / 60 = (24 h, 52 m,) = (1 d, 0 h, 52 m), the error in 4,000 years. This trifling error in the Gregorian calendar may be corrected by suppressing the intercalations in the year 4,000, and its multiples, 8,000, 12,000, 16,000, etc., so that it will not amount to a day in 100,000 years. RULE. Divide 100,000 by 4,000 and you will have the number of intercalations suppressed in 100,000 years. Multiply 1 d, 52 m, (that being the error in 4,000 years) by this quotient, and you will have the discrepancy between the Gregorian and solar year for 100,000 years. By this improved method we suppress 25 days, so that the error will only amount to 25 times 52 minutes. SOLUTION. 100,000 / 4,000 x (1 d, 52 m,) = (25d, 21 h, 40 m.) Now, (25d, 21 h, 40 m,) - 25 d = (21 h, 40 m,) the error in 100,000.