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You may now address the problem of distributing leap years within a mard. Rule 2 states that a year evenly divisible by 2,000 should be a leap year. Hence, years 2, 4, 5, and 10 are possible candidates. You should also like the year selected for variation by rule 1 to be exceptional and to fall at the end of the mard, so that the extra day is epagomenal, rather than intercalary. Because there are six leap years within 10 years, you can see that using either the even or the odd years will produce a simple rule for five of the years. Alternation of leap years tends to smooth the average error, preventing it from growing large before the introduction of a correction. Together, these considerations suggest using the 10th year as the exception, producing a distribution of years within a single mard as shown in Table 9.37. |
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You may thus declare a leap year rule for Mars which is actually simpler than the corresponding rule for the Terran Gregorian calendar: |
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Every odd-numbered year is a leap year. |
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Every year evenly divisible by 10 is a leap year. |
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Any year evenly divisible by 2,000 is not a leap year. |
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Note that the first two rules can be computationally collapsed by taking the year modulo 10 and assigning leap years for remainders that are odd or zero. |
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9.13.2.1.6
Months in Quarters |
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The next step is the assignment of months to quarters within a year. All months contain either 55 or 56 days. A quarter contains (56 + 56 + 55 =) 167 days in a common year. The short month should fall at the end of the quarter, since the last quarter occurs |
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| Table 9.37 Distribution of years within a mard. | | Year in Mard | Year Type | Number of Days | | 1 | Leap | 669 | | 2 | Common | 668 | | 3 | Leap | 669 | | 4 | Common | 668 | | 5 | Leap | 669 | | 6 | Common | 668 | | 7 | Leap | 669 | | 8 | Common | 668 | | 9 | Leap | 669 | | 10 | Leap | 669 |
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