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Or we could do metric time (files.catbox.moe)

submitted 7 months ago by Gork@lemm.ee to c/science_memes@mander.xyz

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[-] rockSlayer@lemmy.world 32 points 7 months ago* (last edited 7 months ago)

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.

[-] Flipper@feddit.de 12 points 7 months ago

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.

[-] LeftHandedWave@lemm.ee 4 points 7 months ago

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

[-] felbane@lemmy.world 17 points 7 months ago* (last edited 7 months ago)

Firstber
Secondary
Thurd
Quadtober
Cincondary
Sextember
Septober
Octuary
Nonuary
Tenber
Postenber
Expostenber
June

[-] Gork@lemm.ee 5 points 7 months ago

Sextember

Nice.

[-] BlackRoseAmongThorns@slrpnk.net 3 points 7 months ago

Wheezed at this, thanks

[-] lugal@sopuli.xyz 1 points 7 months ago

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

[-] ElderWendigo@sh.itjust.works 13 points 7 months ago

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.

[-] lugal@sopuli.xyz 2 points 7 months ago

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

[-] ElderWendigo@sh.itjust.works 2 points 7 months ago* (last edited 7 months ago)

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.

[-] lugal@sopuli.xyz -1 points 7 months ago

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

[-] ElderWendigo@sh.itjust.works 2 points 7 months ago

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.

[-] lugal@sopuli.xyz 1 points 7 months ago

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.

[-] ElderWendigo@sh.itjust.works 1 points 7 months ago

Yes, I understood. I still disagree for the reasons in all of my previous comments.

[-] lugal@sopuli.xyz 0 points 7 months ago

Let's agree to disagree

[-] grue@lemmy.world 8 points 7 months ago

Look, short of changing Earth's orbit, something's not gonna line up no matter what you do. Extra-weekly days are as good a compromise as any in my book.