h a l f b a k e r yThere goes my teleportation concept.
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2 year calendar
print next year's calendar on the back of this year's calendar. | |
So it's getting toward the time that I'll need a new
calendar. I notice that the reverse side of my current
calendar is blank. This is inefficient. Simply print the
following year on this, we know about the dates and so
on, they won't be a surprise. Done, the world can now
expend 50% less
on calendars.
Perpetual calendar
http://en.wikipedia...General_information
14 plus an index, in answer to [MaxwellBuchanan]'s wondering-aloud. [tatterdemalion, Dec 26 2011]
[link]
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I did this, more or less, when i produced the Martian calendar, but it covered a bit less than two of your Earth years so it wasn't quite the same. |
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the backs of *my* calendar are blank. it would halve the work but halve the pleasure of putting it together each year. btw I recycle my Russell Brand calendar every year - the dates are inconsequential :) |
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If you had a twenty-eight year calendar you could
reuse it until next century. |
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Surely one only needs a maximum of 14 calendars?
(The year has to start on one of seven days of the
week, and it's either a leap-year or not.) |
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Yes, but they don't repeat reliably for twenty-eight years, so deciding which one to use is not a no-brainer. You get three a day later, then one partly two days later, then another three, and the two-day shift is on the leap day. Binding them together would help for that reason. |
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Surely all you need to do is wake up on January 1st,
see what day of the week it is, and check if it's a
leap year? |
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So, one only need purchase 28 calendars and be set forever. |
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But how will you tell what year it is ? |
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I realise that seasons encourage people to count in years but i tend to count in days and convert. More specifically, days elapsed since seventeenth July, 'seventy-five, because that was the day i wrote my first diary entry (which are numbered). |
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siderealy, two years is 27 moon revolutions or 729
days which is 3x3x3x3x3x3 which would make the
two years easily divisible into 3 sets of 3 sets of 3,
so why not divide the two year calendar into 3
"years" each of which would be 9 months long, and
each 9-month "year" would contain 3 "seasons",
each of which would be 3 months long, and each
27 day month would contain 3 weeks of 9 days
each and each 9-day week would be split into 3 3-
day sets, and each day would be split into 3
sections and each section, 3 hours, and each hour
3 minutes etc? |
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