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tdn
01-21-2005, 12:42 PM
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?

nivlac
01-21-2005, 12:48 PM
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.

chrisk
01-21-2005, 12:51 PM
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.

CJJ*
01-21-2005, 12:52 PM
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?

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.

chrisk
01-21-2005, 12:54 PM
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.

Quercus
01-21-2005, 12:57 PM
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.

chrisk
01-21-2005, 01:10 PM
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

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 :D )

aahala
01-21-2005, 01:14 PM
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.

tdn
01-21-2005, 01:33 PM
Thanks, everyone. That helps.

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.

Bytegeist
01-21-2005, 04:59 PM
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)

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.

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