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 × 60) ÷ .373 = 3,861, nearly; hence the error would amount to only one day in 3,861 years.

(.373 × 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 × (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.