# Verify my math, please? **URL:** \ **Category:** Factual Questions\ **Created:** [January 21, 2005, 6:42pm UTC](https://boards.straightdope.com/t/verify-my-math-please/285947 "2005-01-21T18:42:04Z")\ **Posts on this page:** 10\ **Page:** 1
**Author:** ![tdn](https://avatars.discourse-cdn.com/v4/letter/t/94ad74/32.png) [@tdn](https://boards.straightdope.com/u/tdn)\ **Post date:** [January 21, 2005, 6:42pm UTC](https://boards.straightdope.com/t/verify-my-math-please/285947/1 "2005-01-21T18:42:04Z")
I feel kind of stupid asking this, but I want to make sure I get it right. I want to calculate the length of the average year. Any particular year needn’t be spot on, but an accumulation of years must approach total accuaracy. A year has 365 days. If the year is divisible by 4, we add a day. If it is divisible by 100, we don’t. But if it’s divisible by 400, it is again. So over the course of a handful of years, the average length is 365 plus .25. Over the course of millenia, would it be 365 plus .25 minus .01 plus .0025? That is, 365.2425? ---
**Author:** ![nivlac](https://avatars.discourse-cdn.com/v4/letter/n/3bc359/32.png) [@nivlac](https://boards.straightdope.com/u/nivlac)\ **Post date:** [January 21, 2005, 6:48pm UTC](https://boards.straightdope.com/t/verify-my-math-please/285947/2 "2005-01-21T18:48:18Z")
> [@tdn](#): > > I feel kind of stupid asking this, but I want to make sure I get it right. > > I want to calculate the length of the average year. Any particular year needn’t be spot on, but an accumulation of years must approach total accuaracy. > > A year has 365 days. If the year is divisible by 4, we add a day. If it is divisible by 100, we don’t. But if it’s divisible by 400, it is again. So over the course of a handful of years, the average length is 365 plus .25. Over the course of millenia, would it be 365 plus .25 minus .01 plus .0025? That is, 365.2425? Depends on what you mean by “year”. If you’re talking about a Julian year, it’s 365.25 days per year on average. If by “year” you mean a sidereal year (how long it takes the earth to make one orbit around the sun), it’s 365.256363051 days. ---
**Author:** ![chrisk](https://avatars.discourse-cdn.com/v4/letter/c/6de8d8/32.png) [@chrisk](https://boards.straightdope.com/u/chrisk)\ **Post date:** [January 21, 2005, 6:51pm UTC](https://boards.straightdope.com/t/verify-my-math-please/285947/3 "2005-01-21T18:51:15Z")
> [@tdn](#): > > I feel kind of stupid asking this, but I want to make sure I get it right. > > I want to calculate the length of the average year. Any particular year needn’t be spot on, but an accumulation of years must approach total accuaracy. > > A year has 365 days. If the year is divisible by 4, we add a day. If it is divisible by 100, we don’t. But if it’s divisible by 400, it is again. So over the course of a handful of years, the average length is 365 plus .25. Over the course of millenia, would it be 365 plus .25 minus .01 plus .0025? That is, 365.2425? Yes, I believe that’s correct, and interestingly, it’s nearly a complete reversal of the way the Gregorian calendar was devised, the way I remember reading it. That is, some bright astronomer figured out that the number of solar days in a seasonal year was approximately 365.2425 , and the gregorian calendar that the western world now observes, (leap year every 4 years, except for century years where the century is NOT a multiple of 4) was derived to match that figure. ---
**Author:** ![CJJ](https://avatars.discourse-cdn.com/v4/letter/c/ecc23a/32.png) [@CJJ](https://boards.straightdope.com/u/CJJ)\ **Post date:** [January 21, 2005, 6:52pm UTC](https://boards.straightdope.com/t/verify-my-math-please/285947/4 "2005-01-21T18:52:32Z")
> [@tdn](#): > > I feel kind of stupid asking this, but I want to make sure I get it right. > > I want to calculate the length of the average year. Any particular year needn’t be spot on, but an accumulation of years must approach total accuaracy. > > A year has 365 days. If the year is divisible by 4, we add a day. If it is divisible by 100, we don’t. But if it’s divisible by 400, it is again. So over the course of a handful of years, the average length is 365 plus .25. Over the course of millenia, would it be 365 plus .25 minus .01 plus .0025? That is, 365.2425? Your math is correct. Another method: Take the interval of 400 years. Each Gregorian year has 365 days, but there are 97 of them that have a leap day (100 years divisible by 4, throw 3 of these out since they end in 00 but are not divisible by 400). So the 365 days of a standard year are augmented by 97/400 days on average, or .2425. So the average length of a Gregorian year over millennia is 365.2425. ---
**Author:** ![chrisk](https://avatars.discourse-cdn.com/v4/letter/c/6de8d8/32.png) [@chrisk](https://boards.straightdope.com/u/chrisk)\ **Post date:** [January 21, 2005, 6:54pm UTC](https://boards.straightdope.com/t/verify-my-math-please/285947/5 "2005-01-21T18:54:14Z")
> [@nivlac](#): > > Depends on what you mean by “year”. If you’re talking about a Julian year, it’s 365.25 days per year on average. If by “year” you mean a sidereal year (how long it takes the earth to make one orbit around the sun), it’s 365.256363051 days. What exactly do you mean by a “sidereal year?” A seasonal year, unless I’m very confused, is the same as sidereal… the axis of the earth doesn’t precess enough to cause a meaningful difference. On the other hand, the sidereal day is definitely different from a solar day, and so the year can be a different number of ‘days’ based on the precise day you are measuring it with. ---
**Author:** ![Quercus](https://avatars.discourse-cdn.com/v4/letter/q/7ab992/32.png) [@Quercus](https://boards.straightdope.com/u/Quercus)\ **Post date:** [January 21, 2005, 6:57pm UTC](https://boards.straightdope.com/t/verify-my-math-please/285947/6 "2005-01-21T18:57:04Z")
Your math is correct – that is is jibes with the Julian calendar. The true length of the Earth year is a little different, and you could find that with some Googling. I found > [@](#): > > On average, the tropical year now lasts 365 days, 5 hours, 48 minutes and 45.20 seconds, and decreases in length by 0.53 seconds per century. I believe that what you’re interested in is the tropical year, which is how long it takes someone on Earth to see the sun reach the same point. ---
**Author:** ![chrisk](https://avatars.discourse-cdn.com/v4/letter/c/6de8d8/32.png) [@chrisk](https://boards.straightdope.com/u/chrisk)\ **Post date:** [January 21, 2005, 7:10pm UTC](https://boards.straightdope.com/t/verify-my-math-please/285947/7 "2005-01-21T19:10:07Z")
> [@Quercus](#): > > Your math is correct – that is is jibes with the Julian calendar. Actually, the [Julian Calendar](http://en.wikipedia.org/wiki/Julian_calendar) has no adjustments every 100 years… so it has an average year length of 365.25 days > [@wikipedia](#): > > However with this scheme too many leap days are added with respect to the astronomical seasons, which on average occur earlier in the calendar by about 11 minutes per year. It is said that Caesar was aware of the discrepancy, but **felt it was of little importance.** In the 16th century the Gregorian calendar reform was introduced to improve its accuracy with respect to the time of the vernal equinox and the synodic month (for Easter). (Emphasis mine… oh mighty Caesar, how little you knew 😃 ) ---
**Author:** ![aahala](https://avatars.discourse-cdn.com/v4/letter/a/a88e4f/32.png) [@aahala](https://boards.straightdope.com/u/aahala)\ **Post date:** [January 21, 2005, 7:14pm UTC](https://boards.straightdope.com/t/verify-my-math-please/285947/8 "2005-01-21T19:14:48Z")
I agree with the math. Now I have a question. When calculating an average, under what circumstances does it matter whether you report it accurate to 5 or 50 decimal points? Averages by their very nature are “fuzzy” things. ---
**Author:** ![tdn](https://avatars.discourse-cdn.com/v4/letter/t/94ad74/32.png) [@tdn](https://boards.straightdope.com/u/tdn)\ **Post date:** [January 21, 2005, 7:33pm UTC](https://boards.straightdope.com/t/verify-my-math-please/285947/9 "2005-01-21T19:33:00Z")
Thanks, everyone. That helps. > [@Quercus](#): > > I believe that what you’re interested in is the tropical year, which is how long it takes someone on Earth to see the sun reach the same point. Actually, I’m not, I’m interested in the calendar year. What I need to do is approximate the difference between two dates and slap a label on the later one. If a patient has an operation, then comes in years later for a followup, I want to know whether to label it as a “4 year followup” or a “5 year followup”, for example. For this, I need to figure out which year the followup date is closer to, and so I need a constant to define a year, based on days. Since most patients end their relationship with us after the 5 year followup, 365.25 is more than accurate enough, in fact 365 is probably close enough. But I’m a geek, so I want to use the best number that I can reasonably get my hands on. However, while the basic answer appeals to the programmer in me, the other stuff appeals to the astronomer in me, so it’s all good. ---
**Author:** ![Bytegeist](https://avatars.discourse-cdn.com/v4/letter/b/9dc877/32.png) [@Bytegeist](https://boards.straightdope.com/u/Bytegeist)\ **Post date:** [January 21, 2005, 10:59pm UTC](https://boards.straightdope.com/t/verify-my-math-please/285947/10 "2005-01-21T22:59:52Z")
A bit more discussion of the various years (there are at least three definitions useful to astronomers) can be found [here.](http://www.maa.mhn.de/Scholar/calendar.html) > [@tdn](#): > > Actually, I’m not [interested in the tropical year], I’m interested in the calendar year. Fair enough — though the Gregorian calendar we use is designed to track the tropical year, on average. Obviously it doesn’t quite hit it, leaving a residual slippage of about one day for every 3300 years.