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| weekcountingproblem [2026/02/13 02:14] – thehomeland | weekcountingproblem [2026/02/13 02:42] (current) – [Possible Solutions] thehomeland | ||
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| | 9m | 10m | 11m | 12m | 12m1w | | | 9m | 10m | 11m | 12m | 12m1w | | ||
| - | The question remains: | + | ==== The question remains: |
| + | ---- | ||
| + | 1. Need we bother, with ensuring each year comes up to 52 counts (since counts are not tallied in the actual process), and just leave it at the incompleteness of 52? The simple answer to that is no, and to just move on. | ||
| + | |||
| + | 2. If do need bother, what solution should we apply, to ensure each of the 52 weeks is accounted for, but without interrupting the normal flow of the process? | ||
| + | |||
| + | ==== Possible Solutions ==== | ||
| + | ---- | ||
| + | |||
| + | Let's explore adding 1 leap weeks every 6 months: insert a 6m0w leap-week every 6 months, as a separate count from the normal 6m, so that we go 5m2w, 5m3w, 6m, 6m0w, 6m1w, 6m2w. That would show up as: | ||
| + | |||
| + | ^ 1 count ^ 5 counts ^ 9 counts ^ 13 counts ^ 17 counts ^ | ||
| + | | | 1w | 1m1w | 2m1w | 3m1w | | ||
| + | | | 2w | 1m2w | 2m2w | 3m2w | | ||
| + | | | 3w | 1m3w | 2m3w | 3m3w | | ||
| + | | | 1m | 2m | 3m | 4m | | ||
| + | | 0m0w | | | | | | ||
| + | |||
| + | ^ 21 counts ^ 26 counts ^ 30 counts ^ 34 counts ^ | ||
| + | | 4m1w | 5m1w | 6m1w | 7m1w | | ||
| + | | 4m2w | 5m2w | 6m2w | 7m2w | | ||
| + | | 4m3w | 5m3w | 6m3w | 7m3w | | ||
| + | | 5m | 6m | 7m | 8m | | ||
| + | | | 6m0w | | | | ||
| + | |||
| + | |||
| + | ^ 38 counts ^ 42 counts ^ 46 counts ^ 51 counts ^ 52 counts ^ | ||
| + | | 8m1w | 9m1w | 10m1w | 11m1w | 1y1w | | ||
| + | | 8m2w | 9m2w | 10m2w | 11m2w | | | ||
| + | | 8m3w | 9m3w | 10m3w | 11m3w | | | ||
| + | | 9m | 10m | 11m | 1y | | | ||
| + | | | | | 1y0w | | | ||
| + | |||
| + | This seems like a viable solution, but wouldn' | ||