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Or we could do metric time (files.catbox.moe)
submitted 2 years ago by to c/science_memes@mander.xyz
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[–] 77 points 2 years ago (3 children)

Bad news: this has nothing to do with timezones, leap days nor daylight saving time. Honestly, leap days would be worse because they wouldn't be part of the 7 day week

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  • [–] 30 points 2 years ago* (last edited 2 years ago) (2 children)

    It's accounted for just like any other leap year, add it to the end of a month as a universal holiday. Most calendar models make it July 29. It's also worth noting that this is actually 364 days, and a single day at the end of the year is a universal holiday.

    Edit: I think leap years should be at the end of the year too for simplicity.

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  • [–] 12 points 2 years ago (1 child)

    That would just be new year. I've already have a list ready for how to name all the months, so we don't fuck it up like September being the 9. Month again.

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  • [–] 4 points 2 years ago (1 child)

    Ooh, tell me what the names would be! Don't leave me hanging. I HATE that September - December are all off.

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  • [–] 0 points 2 years ago (3 children)

    Which breaks "day of week = day modulo 7" if every month starts on Monday and not every month has the same number of days

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  • [–] 14 points 2 years ago (1 child)

    In this scheme, new years day and leap days are not any day of the week or part of any month. They exist outside of the regular calendar as obvious and explicit resets to the remainder problem.

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  • [–] 2 points 2 years ago (1 child)

    My point exactly. So the programmer who commented above me is wrong in saying it makes it easier for them

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  • [–] 2 points 2 years ago* (last edited 2 years ago) (1 child)

    No, still easier. They are still part of the year, so you can just count them, and the logic is still easier than the mess we currently have. If you really feel the need to you can call new years day the zeroth day in the zeroth month, the day of the week is Holiday, and periodically the zeroth month has one extra Holiday.

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  • [–] -1 points 2 years ago (1 child)

    Computers store the date as "days after January 1st 1970". So you have a huge number, divide it with 7 and get the day of the week. If there are days that don't belong to any week, you have to calculate January 1st of that year and substrate it in addition to the steps above. I don't say it's not manageable, but it's not easier

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  • [–] 2 points 2 years ago (1 child)

    They store the number of seconds since the epoch of 1970, but you're always going to have leap days and even leap seconds. Even if you changed the definition of a second to match the current length of a year, it would be off again relatively soon and you'd need leap seconds again. It's NEVER going to be as simple as you seem to think it should be. Chaos and complexity is inherent in the whole system.

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  • [–] 1 point 2 years ago (1 child)

    I never said it was simple. The comment above me was "oh, this makes it much easier" and I was like "it's not really getting easier". That's all I said.

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  • [–] 1 point 2 years ago (1 child)

    Leap day gets it's own name outside of saturday through sunday. It's an all awesome holiday.

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  • [–] 0 points 2 years ago (1 child)

    ... which fucks with the way the day of the week is calculated by computers as I already explained others

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