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This is easily done, since 6/10 = 0.6. Thus, rule 1 is that a basic unit of 10 years (as opposed to four on Earth) will be used, with leap years spread among six of them, in a pattern somewhat like the Islamic or Jewish rule. Four years of every 10 must be common years (668 days) and six must be leap years (669 days). The group of 10 years is called a Martian decade, hereafter abbreviated to mard. |
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With this system, there are (4668 + 6669 =) 6,686 days in 10 years, giving an average of 668.6 days per year. The error is thus (0.60 0.5994731 =) 0.0005269 days per year, which is already much better than the Gregorian calendar. The next question that naturally arises concerns the distribution of these years within a mard. Postpone this decision until you have looked at the remaining error and determined the next rules. These rules may introduce certain patterns that can be more easily implemented by a clever assignment of leap years within a mard. Using the basic system, determine the remaining error, that is (0.60 0.5994731 =) 0.0005269 days per year. This error will accumulate to 5.269 days in 10,000 years, or one day every 1,897.89333 years, or about one day every 2,000 years. This result suggests rule 2 the year is not a leap year every 2,000 years. Note that this rule influences the choice for distributing leap years within one mard, since you will want a year evenly divisible by 2,000 to be a leap year by rule 1, so that rule 2 can make it an exception. |
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Hence, the second rule is simple drop a day every 2,000 years. You can do that by declaring years evenly divisible by 2,000 to be common years. Looking at the number of days within 2,000 years, this rule then gives: |
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| Mard# | 1 | 2 | 3 | . . . | 199 | 200 | | No. Days | 6,686 | 6,686 | 6,686 | . . . | 6,686 | 6,685 | | Years | 110 | 1120 | 2130 | . . . | 19811990 | 19912000 |
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The accuracy of the calendar is determined by looking at the number of days in one period of 2,000 Martian years. This ''Martian bimillennium" recurs frequently enough in this discussion to warrant a name, which will be the contraction marbimill. Each marbimill contains 200 mards. |
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The first marbimill contains (1996686 + 6685 =) 1,337,199 days, giving an average of 1,337,199/2,000 = 668.5995 days per year. This result gives us an error of 668.5995 668.5994731 = 0.0000269 days, or one day every 37,174.72119 years. This error is well below the threshold of one day in 32,000 years. There is no advantage to pursuing accuracy beyond this point for two reasons: |
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The year is kept as a type int in the DATE_INFO structure, so the maximum positive value that can be represented is INT_MAX, or 32,767. |
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The rotation of planets varies in ways not completely understood at this time. As a consequence, the actual length of a day contains error terms that become significant after about 30,000 years. This conclusion applies to Earth. The Martian value probably differs. |
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