4 ms·
Simple Decimal Calendar
- Mindey 3y agoUnix days make for a good decimal calendar. Simply divide Unix seconds by 86400, and you're done. While there are several decimal calendar attempts available online, this approach follows Occam's Razor principle by relying on basic division and digit grouping without the need for additional inventions, and yet, has pretty much all the benefits one would want.
- JoeAltmaier 3y agoDays? What use are 'days' on Mars? A different approach to time is desirable, instead of being loosely related to celestial mechanics. How about: a shorter second. 100,000 seconds in a 'day'. 1M seconds in a 'week'. and so on. On a space station they sure wouldn't mind the difference. They'd talk about 'see you in a klik' meaning 1000 seconds, or 'thatll take about a megasec' meaning a week. 86400? Give me a break.
- Mindey 3y ago> How about: a shorter second. 100,000 seconds in a 'day'. 1M seconds in a 'week'. and so on. Yes, that's exactly what you get when dividing Unix secs by 86400 you get 100,000-second-long days, useful on Earth, for now. :) What would be a non-celestial day?
- JoeAltmaier 3y agoA day? Nothing. It doesn't exist. A hundred thousand seconds perhaps would be a useful measure; perhaps not.
- Mindey 3y agoNormalizing Unix seconds by the number of seconds in a day, results in Unix day, which has convenient properties, such as 5th digit after decimal point corresponding to decimal second (0.864 length of traditional second). So, dividing UNIX_SECOND by 86400, we get a decimal Unix day. For example, if UNIX_SECOND = 1688897360, then UNIX DAY = 19547.42315, Displaying this number with time formatting, makes it extremely easy for our time-trained eyes to interpret: 19,547 4:23:15, meaning: 19th Unix decimal year 547th day of the year (5-dmonth, 4-dweek, 7-dweekday) 4:23:15 decimal time (4-dhour, 23-dminute, 15-dsecond)