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.
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.
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
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.
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.
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.
Which breaks “day of week = day modulo 7” if every month starts on Monday and not every month has the same number of days
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.
My point exactly. So the programmer who commented above me is wrong in saying it makes it easier for them
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.
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
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.
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.
Yes, I understood. I still disagree for the reasons in all of my previous comments.
Let’s agree to disagree
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.
There is also a technological solution, I knew it
Leap day and new year day are supposed to not be a week day in this system
My point exactly. So the programmer who commented above me is wrong in saying it makes it easier for them