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I have this idea of keeping track of your age in decimals. I figured out the dates the other day of when I turn one-tenth years older. So on July 2nd, at noon, I will be 19.1 years old and on September 13th, I will be 19.3 years old. Not sure if I will have the times EXACTLY correct because of leap years or the actual length of time there is in a year, but I think its pretty neat.
1 year ago · 👍 cobradile94, krixano
@krixano not that it's super important, but I checked and the lubridate package (correctly) identifies 1900 as a non-leap-year. · 1 year ago
Yeah, it is interesting. I suppose it wouldn't matter as much for ages exactly, since nobody (or barely anybody) lives more than 128 years. But if we were to use the calculation for the age of long-lasting things, like countries for example, then it would matter more. · 1 year ago
@krixano you are absolutely correct. Did some quick digging into the exacts. If were being as acurate as possible, the decimal expands out to:
365.2421891 days, in more human readable terms: 365 days, 5 hours, 48 minutes, 45 seconds, and 138 milliseconds.
I also found it interesting how long it takes for a calendar to drift a day depending on how many decimals we allow.
365 day calander drifts 1 day in just 4 years
365.25 drifts 1 day in 128 years
365.2425 drifts 1 day in 3,216 years · 1 year ago
@smokey I believe 365.25 is for the Julian calendar, but the Gregorian calendar made the leap year calculations more complex, so the number is more like 365.2425 · 1 year ago
@degrowther2 Yeah, likely the implementation of those functions uses julian day calculations, which are pretty simple to do. I believe you can get the calculations for turning a gregorian date into a julian day via the wikipedia page for "Julian Day". I also have the calculations as well, if people can't find them. There's also the book recommended by the NOAA called "Astronomical Algorithms" that has this as one of its first chapters. · 1 year ago
I use the R programming language for my day job, and will often use the lubridate package to count days between various events. For instance, I am approximately 16,000 days old. · 1 year ago
lets see if my numbers check out
1 year / ~365.25 days (add in .25 to factor in the leap year)
1/10th of 365.25 is 36.525 days
Start at the date of your birthday, and each 36 and a little over a half days mark a new 10th incriment to measure. · 1 year ago
Ok, so I just did this experiment, it turns out that this very second I am 24.3509301532 years old :D · 1 year ago
You can probably get more accurate results if you turned your birthday and the curent day into Julian days, and then subtract two of them to get the number of days between, and then just divide by 365.2425 (for Gregorian Calendar years). This would also account for leap years, and would work for other calendars as well, and would even be accurate to the very second of a day :D
I actually have the ability to do something like this with one of my programs, should be a simple experiment. · 1 year ago