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Table 9.39 shows the accumulated error for a 100-dar period. The error figure represents the cumulative error at the end of each dar between the actual length of a dar and the official clock. |
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The second case is that of adding one second every four dars. In this case, the rule adds (100/4 =) 25 seconds in 100 dars, which is two seconds more than required. Hence, a second rule must be employed to effectively remove two leap seconds from the count. This may be accomplished by the following rule set: |
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1.) Every fourth dar, add one second at the end of the dar. |
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2.) On dar 50, subtract one second. |
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3.) On dar 100, do not add one second. |
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Because 100 is evenly divisible by four, dar 100 adds one second by rule 1. Preventing this addition by rule 3 is the same as allowing the addition and then reversing it with a subtraction. So, an equivalent set for rules 2 and 3 would be: |
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| Table 9.39 Accumulated error for 100-dar period: case 1. | | Dar Number | Accumulated Error(s) | | 1 | 0.23 | | 2 | 0.46 | | 3 | 0.69 | | 4 | 0.92 | | 5 | 0.15 (1.15 -> 0.15 after leap second added) | | 10 | 0.30 | | 15 | 0.45 | | 20 | 0.60 | | 25 | 0.75 | | 30 | 0.90 | | 35 | 0.05 (1.05 -> 0.05 after two leap seconds added) | | 40 | 0.20 | | 45 | 0.35 | | 50 | 0.50 | | 55 | 0.65 | | 60 | 0.80 | | 65 | 0.95 | | 70 | 0.10 (1.10 -> 0.10 after two leap seconds added) | | 75 | 0.25 | | 80 | 0.40 | | 85 | 0.55 | | 90 | 0.70 | | 95 | 0.85 | | 100 | 0.00 (1.00 -> 0.00 after two leap seconds added) |
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