How to Scientifically Measure How Many Days Are in a Year
Table of Contents
Summary
- The article explores the measurement of the number of days in a year, considering various definitions of a “day” and a “year.”
- It starts with the common knowledge of a year consisting of 365 days, with leap years having 366 days.
- The concept of a “stellar day,” based on Earth’s rotation with respect to fixed stars, is introduced.
- The “sidereal year,” the time it takes for Earth to orbit the Sun with respect to the stars, is discussed, revealing a slightly longer duration than previously thought.
- Differences between a stellar day and a 24-hour day are attributed to the Earth’s elliptical orbit and its effect on solar day length.
- Precession, the Earth’s axis wobbling, is explained, causing long-term changes in star positions and the Sun’s location.
- The article summarizes various measures of a year, including the tropical year and anomalistic year, and their relationship to Earth’s motions.
How many days are there in a year?
Ask pretty much anyone and they will tell you 365, except for leaps years which have 366. In addition, most people can tell you that a year is the amount of time that it takes for the Earth to go once around the Sun, and a day is the amount of time it takes for the Earth to rotate on its axis once. (I say “most”, because, in a survey done too long ago, only 74% of Americans surveyed even knew that the Earth went around the Sun.) Given the fact that leap years occur every four years (with exceptions), it turns out that the Earth rotates ~365.25 times for every trip it makes around the Sun, so you could say that there are ~365.25 days in a year. (The Julian calendar assumed that there were 365.25 days in the year and added a leap year every four years to account for that extra ¼ of a day. And while the Julian calendar has since been replaced with a more accurate one, astronomers still use Julian dates in many of their calculations. In fact, the light-year is the distance light travels in a Julian year.)
However, the above definitions of “day” and “year” are a little vague. One trip around the Sun and one rotation on the Earth’s axis, with respect to what? The reference often used by astronomers is that of the “fixed” stars. If we measure one rotation of the Earth with respect to the stars we call it a “stellar day”. And using the same reference, one trip of the Earth around the Sun with respect to the stars is called a “sidereal year”, from the Latin word “sidus” meaning star. (from this you would think that a rotation of the Earth with respect to the stars would be called a sidereal day, but oddly, a sidereal day is measured with respect to a different reference which we will deal with later on.)
If you measure the Earth’s rotation with respect to the stars you get 86164.098903691 seconds per rotation, and if you do the same for the Earth orbit around the Sun, you get 31558149.76… sec. This means that, with respect to the stars, the Earth rotates 366.25636623 times while it orbits the Sun once. This means that there are 366.25636623 stellar days to the sidereal year This is at least one whole “day” more than the 365.25 we came up with earlier. So where does this difference come from?
You may have noticed that the length of a stellar day differs from what we normally consider as a day of 24 hrs, with 60 min hrs and 60 sec minutes which works out to 86400 sec, and is almost 4 minutes longer than the stellar day. This is because our 24hr clock is based on the “Solar day”, or the time it takes for the Sun to go from noon on one day to noon of the next, and this time period is affected by the Earth’s orbit around the Sun.
In the time it takes the Earth to make one rotation with respect to the stars, the Earth travels a bit under 1 degree around the Sun. So a spot that starts directly under the Sun will not be so 1 stellar day later.
This means that the Earth must continue rotating a bit longer (~ 4 min) to finally return that spot directly under the Sun. So it turns out that 1 solar day equals 1.002737803 stellar days.
( The length of the solar day worked out above is for the “mean” solar day. The Earth does not orbit the Sun in a perfect circle but in an ellipse. Because of this, its orbital and angular speed around the Sun varies over the course of its orbit. It travels faster near perihelion and slower at aphelion. This makes the solar days when the Earth is near perihelion longer than those when we are near aphelion. At present, perihelion occurs in December, so this is when the longest solar days occur. So for those of you in the Northern hemisphere who feel that Winter days seem to drag on while Summer days fly by, you are not entirely incorrect. This will not always be the case, however, as we will learn later.)
One of the side effects of this difference between the stellar and solar day is that the stars that we see in the night sky change over the course of a year. The Winter night sky will be different from the Summer night sky. In addition, if we could see the stars during the day, the constellation that we would see the Sun in also changes over the course of the year. (This is the basis of astrology. The Sun passes through twelve constellations which are the “signs” of the Zodiac in a year. The constellation in which the Sun is in on your birth date is your Sun sign.)
Sun’s position in Taurus on May 21, 2015
So now we can work out that there are 365.2563662 mean solar days to a sidereal year, which is slightly more than the 365.25 days per year we estimated earlier.
This means that if you wished to create a calendar that kept step with the sidereal year, you could start with the Julian calendar and then “fudge” it to make it more accurate
The Gregorian calendar, which is now used in most of the Western world is a result of such fudging.
The Gregorian calendar is accurate to within 1 day in several thousand years when it comes to keeping the calendar in step with the seasons (The aim of most calendars). To accomplish this, leap days were eliminated on century years that were not evenly divisible by 400. (1900 was not a leap year, but 2000 was.) So every 400 years there are 97 leap days added for a total of 146097 days vs 100 leap days for a total of 146100 with the Julian calendar.
But hold on. With 365.2563662 days to a sidereal year, in 400 sidereal years, you end up with 146102.5465 days or over 5 days more than accounted for by the Gregorian calendar and 2.5465 days more than accounted for by the Julian calendar. This means if we want to keep step with the sidereal year, then the Julian calendar is a better fit than the Gregorian calendar and in order to make it an even better fit, we would have to add more leaps years not fewer.
But the Gregorian year is a closer fit to the change of the seasons, why is it a worse fit to the sidereal year?
The reason is that the sidereal year and the seasons do not keep in step.
To understand why this is, we need to look into what causes the seasons, and this is the tilt of the Earth’s axis with respect to the Sun.
When the Northern hemisphere leans toward the Sun we have Summer, and when the Northern hemisphere leans away from it we have Winter. The moment the Northern hemisphere is inclined at its greatest to the Sun is called the Summer solstice and when it is inclined away by the greatest amount, it is called the Winter solstice. The Autumn and Spring (or Vernal) equinoxes are when the tilt is neither towards nor away.
The Northernmost line of latitude where the Sun can be directly overhead (during the Summer Solstice) is called the Tropic of Cancer. (so-called because astrologically speaking, the Sun sign at this time is Cancer.) The Southernmost line of latitude where the Sun can be directly overhead during the Winter solstice is the Tropic of Capricorn, so named for the same reason.)
The time period between Vernal equinoxes is called a tropical year. Because during this period the Sun goes from being directly overhead the equator at noon to overhead the Tropic of Cancer, to over the equator, to over the Tropic of Capricorn, and finally back to the equator again. This is also the seasonal year and so the Gregorian calendar is designed to fit the tropical year.
Now if this were all there was to it, and the Earth’s axis always pointed in the same direction with respect to the stars, the seasons would keep in step with the Earth’s trip around the Sun and the tropical year and sidereal year would be equal in length. But as we saw above, this isn’t the case.
The truth is that the Earth’s axis doesn’t stay fixed with respect to the stars; It “wobbles” like a spinning top. This wobble is called precession.
A spinning top precesses because it is trying to fall over. Gravity pulls on it and applies torque to the axis of rotation. The top, in turn, tries to preserve angular momentum, and the torque is converted to a precession of the axis.
With the Earth, gravity is the culprit also, Both the Sun and Moon (and the other planets to a certain degree) exert a gravitational effect on the Earth, each pulling on the far side a little less than the near side. If the Earth was perfectly spherical, this would not matter, but the Earth is a slightly oblate spheroid. The effect of this gravitational difference across the Earth is that both the Moon and Sun exert a torque on the Earth which tries to align the equatorial bulge with the planes of the respective orbits. The combined torque exerted by both these bodies has the same effect as the gravity trying to topple over the spinning top and causes the spinning Earth to precess.
The resulting precession causes the Earth’s axis to trace out one complete circle with respect to the stars every 25770 years. This means if you started with the Northern hemisphere leaning towards the Sun, after the 12885 sidereal years, the Earth’s axis will also have precessed ~180 degrees with respect to the stars and you will have the Northern hemisphere leaning away from Sun after those 12885 sidereal years.
Thus with a calendar based on the sidereal year, the seasons would slowly drift with respect to the calendar, and you’d eventually have Summer weather occurring in December. The end result is that the tropical year is slightly shorter than the sidereal year, being 365.24219 mean solar days long vs. 365.2563662 mean solar days to a sidereal year. 365.2563662 times 400 equals 146096.876, which is very close to the 146097 days in the Gregorian calendar for the same period, which is why it is such a better fit.
This difference of 0.0141762 days per year also leads to what is known as the precession of the equinoxes. If we compare the relative position of the Earth at the Vernal equinox with respect to the stars, we find that it changes over time.
(And this is why we have both a sidereal and stellar day as I mentioned earlier; the sidereal day is measured with respect to the Vernal equinox and the stellar with respect to the stars, making the sidereal day slightly shorter than the Stellar day)
Further, this means that not only does the night sky change day by day, it also changes over the years. On any given date, the night sky changes slightly from year to year. It isn’t much and isn’t really noticeable from one year to the next, but over the course of an average person’s lifetime it works out to be a shift of a bit over twice the width of the full Moon.
This also means that the position of the Sun with respect to the stars on a given date as seen from the Earth also changes over the years. In the time since the dates for the astrological signs were established, the Sun has moved an entire Zodiac sign. Thus, during the Summer solstice this year, instead of just entering Cancer as the astrological dates suggest, the Sun is just leaving Taurus and entering Gemini. (So if the Tropic of Cancer had been named today, it would have likely been called the Tropic of Gemini and the Tropic of Capricorn would be the Tropic of Sagittarius).
So if you really wanted to keep the astrological signs in step with the constellation the that Sun is in on a particular day, you would need to shift the dates of the astrological sign by 1 day every 68 years or so. ( Another indication of the silliness of astrology, as it does not even keep consistent with its original premise that the sign(constellation) the Sun is in on the day you were born determines your personality.)
So now we can finally answer the question: How many days are in a year?
Considering the 2 different “years”, sidereal and tropical, and 3 different “days”, stellar, solar, and sidereal a list comparing them all would look like this.
365.24219 solar days to a tropical year
365.256363009 solar days to a sidereal year
366.2421544183 stellar days to a tropical year
366.2421900073 sidereal days to a tropical year
366.2563662303 stellar days to a sidereal year
366.2564018207 sidereal days to a sidereal year.
Which pretty much wraps up the question. Except…
Remember earlier when I mentioned that perihelion of the Earth’s orbit occurred in December and that this made Winter days longer? I also said that this will not always be the case. Here’s why:
I already mentioned that the Earth does not orbit in a perfect circle but in an ellipse, with a perihelion and aphelion. The time it takes for the Earth to go from perihelion to perihelion is an “anomalistic year” (because the angular distance between a planet and its last perihelion is called the anomaly). Because the Earth’s orbit is subject to the gravitational tugs of other bodies in the Solar system, it undergoes a precession so that the perihelion shifts a bit over time with respect to the stars. Thus the time it takes for the Earth to go from perihelion to perihelion is just a bit longer than a sidereal year, which in turn is longer than a Tropical year, making the anomalistic year 365.259636 mean solar days long.
This also means that the perihelion date changes slightly from year to year, or by about 1 day every 57 years.
So for those of you lamenting those long Winter days in the Northern hemisphere, take heart, in 10,000 years or so they will be a bit shorter as the perihelion shifts to the Summer months.
Next: Here Be Dragons.
[QUOTE=”Jim60, post: 5453234, member: 590906″]Has anyone on this forum attempted to count the days in a year using this method, or any other way for that matter,[/QUOTE]
Here’s another way: I plot the mileage my car gets and fit it to a sinusoid plus a constant. The year is 363.2 +/- 1.2 days long.
[QUOTE=”mfb, post: 5451178, member: 405866″]Only if you want to observe the sunrise every day in the year, which is not necessary.[/QUOTE]
You make sound easy?
Has anyone on this forum attempted to count the days in a year using this method, or any other way for that matter, just to see if it could be done?
[QUOTE=”Jim60, post: 5450969, member: 590906″]That would mean you would need a total angle of about 90 degrees clear view of the horizon.[/QUOTE]Only if you want to observe the sunrise every day in the year, which is not necessary.
I should have put UTC after minutes, silly me.
I’ve notice most astronomical events have a Julian Day for a time; it starts from noon January 1st 4713 BC.
For example this years Perihelion occurred on January 2nd 2016 at 22 hours 49 minutes, or 2457390.4506944400 JD (Julian Day).
That’s if you believe that site I mentioned on page 2.
Very interesting topic indeed. Thank you Janus.
Reading that gives one a bit of sympathy for those who allowed the Y2K bug to exist. Doing date and time correctly in software is pretty difficult. Most programmers were not up to dealing with the complexity and they gave up in disgust. Even those who were up to it had multiple definitions to choose from.
One might expect that a standard time/date library would have been developed even in the days before open source. But (at least) two families of versions would be needed, a scientific family, and a human family. The scientific versions (UTC is one such) would be used for stuff like astronomy. (The Insights article reminds us that there are multiple versions of that.) The human versions would be used for stuff like when does the next train arrive, how much to budget for today’s hourly wages (23, 24, or 25 hours?), and how much electric energy will be consumed tomorrow (depends on the day of the week and holidays). Obviously, the human versions would need geographical and cultural instantiations.
Then consider the type conversion problems as real life time/date data collected came from incompatible versions. How many versions of DAYSDIFF(TIMEDATE1,TIMEDATE2) would we need to cover all the combinations of definitions?
Even today, if an open source library exists that is able to deal with all the scientific and human definitions of date/time (past and present), I’m not aware of it.
So thanks again Janus for reminding us that “What time is it?” is a question whose answer we can never find universal agreement. Not in the past; not now; not in the future.
Did you notice on that site that the Solstices, Equinoxes and Perihelion to Perihelion varied quite a lot from the norm?
Is this the Solar Azimuth Angle degrees clockwise from north?
If that’s the case my sunrise on the 20/3/2016 was 6.07am LST, and the Suns azimuth angle at that time was about 88.45 degrees north/east.
On 19/6/2016 the Sunrise will occur at 3.06am LST, and the Suns azimuth angle at that time will be about 40.18 degrees north/east.
On 22/9/2016 the Sunrise will occur at 5.51am LST, and the Suns azimuth angle at that time will be about 88.96 degrees north/east.
On 21/12/2016 the Sunrise will occur at 8.46am LST, and the Suns azimuth angle at that time will be about 134.88 degrees south/east.
On 20/3/2017 the Sunrise will occur at 6.08am LST, and the Suns azimuth angle at that time will be about 89.75 degrees north/east.
So sunrise would move about 40.18 degrees north/east to 134.88 degrees south/east.
That would mean you would need a total angle of about 90 degrees clear view of the horizon.
Is this correct or have got it wrong?
It should not be 365.25.
Independent of that, there are smaller variations, some regular, some not.[QUOTE=”Jim60, post: 5449743, member: 590906″]If we take the Suns position on the horizon as true east on March 20th, how many degrees does the Sun move north and south of east in total over the year?[/QUOTE]It depends on your location. If I did not make a mistake, those are the angles (in degree) for different latitude, with 23.5 degrees tilt of Earth:
66: 157 <- but hard to observe as the sun barely makes it above the horizon at winter solstice where most of that change happens. 60: 106 50: 77 40: 63 30: 55 20: 50 10: 48 0: 47 <- much easier to observe thanks to nearly vertical sunrise/set
Checking how many days from equinox to equinox using this site, I’m puzzled that the time varies from year to year.
[URL]http://aa.usno.navy.mil/data/docs/EarthSeasons.php[/URL]
This year’s equinox fell on March the 20th at 4 hours 30 minutes UTC, and next years equinox falls on March 20th at 10 hours 29 minutes UTC, a difference of 365.25 – 365.249 = 1 minute less than it should be.
Either I can’t do simple maths, or there is something else that I would have to take into account if wanted to find out how many days from equinox to equinox.
Orienting it would be a delicate process, but offhand I would think you could set one up just about anywhere.
Yes you could. The shadow would move steadily clockwise from sunrise till sunset.
Great skill would be needed in setting it up though; the location would have to be ideal.
[QUOTE=”Jim60, post: 5449794, member: 590906″]That equatorial ring looks like it does the same job as the equatorial mount on a telescope, where it compensates for the 23 degrees tilt of the Earth?
I presume you would check the Suns position with that while it’s on the meridian at mid day?[/QUOTE]
No, I think you can check it at any time of the daytime. If oriented correctly, the bottom of the ring looks like it should in shadow for the entirety of the day.
That equatorial ring looks like it does the same job as the equatorial mount on a telescope, where it compensates for the 23 degrees tilt of the Earth?
I presume you would check the Suns position with that while it’s on the meridian at mid day?
If we take the Suns position on the horizon as true east on March 20th, how many degrees does the Sun move north and south of east in total over the year?
The reason for asking this, is because you would need a fairly wide horizon east without any obstructions.
It would also be helpful in designing something to mark the position of each Sun rise.
[QUOTE=”Jim60, post: 5449207, member: 590906″]Would it be easier to start counting the days from March 20th the spring equinox, as the Sun would rise in the east and set in the west on that day?[/QUOTE]There is nothing special about a sunrise exactly east and sunset exactly west – unless you use something like the equatorial ring, but that needs alignment before (and then you could also use it for different days, with different alignment).
A precision of an hour is great, that gives 365.25 within a single year, with “2” as significant figure already. 10 years and you start noting that 365.24 is a better approximation.
[QUOTE=”Jim60, post: 5449207, member: 590906″]Would it be easier to start counting the days from March 20th the spring equinox, as the Sun would rise in the east and set in the west on that day?[/QUOTE]
As far as I know, the sun rises in the east and sets in the west every day.
[QUOTE]
After that day the Sun would start rising more northerly until about June 19th, then after that day the Sun would start to rise more easterly again.
Eventually the Sun would be back to where it started on about September 22nd
After September 22nd the Sun would start to rise more southerly until around December 21st.Then the Sun would return to where it started on March 20th 365.25 days later.
Hope this waffle can be understood.
Have I missed something?[/QUOTE]
Hipparchus and other Greek astronomers used a device called an [B]equatorial ring[/B] to determine when the equinoxes occurred:
[URL]https://en.wikipedia.org/wiki/Equatorial_ring[/URL]
Would it be easier to start counting the days from March 20th the spring equinox, as the Sun would rise in the east and set in the west on that day?
After that day the Sun would start rising more northerly until about June 19th, then after that day the Sun would start to rise more easterly again.
Eventually the Sun would be back to where it started on about September 22nd
After September 22nd the Sun would start to rise more southerly until around December 21st.Then the Sun would return to where it started on March 20th 365.25 days later.
Hope this waffle can be understood.
Have I missed something?
[QUOTE=”Jim60, post: 5447852, member: 590906″]Thanks very much for all the replies.
Has anyone tried timing the Sun or Sirius to get the length of a year?
If anybody can give me some tips, I’ll give it a go myself.
Getting 365.25… Day’s without any modern equipment like an accurate clock and a precision mounted telescope seems impossible, how did the Romans achieve it?[/QUOTE]
It wasn’t the Romans who did. It was the Greeks among others who made the observations necessary.
Hipparchus of Nicaea was the premier astronomer of his time, insisting that Greek astronomy be done just as precisely and meticulously as the Babylonians did.
[URL]https://en.wikipedia.org/wiki/Hipparchus[/URL]
Hipparchus had access only to simple tools and instruments. The length of the year was made by observing the equinoxes.
You don’t need any timing, apart from counting days.
Find some place with a clear view to the horizon. where the sun rises or sets. On a clear day, observe where it does so, note down your own position and the apparent position on the horizon. Ideally, repeat this a few times to get several data points, and to get a feeling how much this changes over time. Always write down how many days passed since the first data point. In about 4 months, the sun will approach the same position for sunset/sunrise again, but this time going in the opposite direction from day to day. In about a year, the sun will approach the same position again – find the day that is closest to the initial position. If you do it right, you should get something between 363 and 367 days if you live not too close to the equator or the poles. Wait another year (but count days), repeat the measurement. By now you can probably narrow it down to +- 1 day. Multiple measurements and interpolation between the days allows to get more precise. Alternatively, do the observations for a few years and you get a much better precision.
The disk of the sun has an apparent diameter of 1/2 degree, if you measure its position with an uncertainty of 1/2 the sun diameter you get 1/4 degree resolution, which should be of the order of the change within a day.
A building or similar tall structure at the horizon allows to make measurements much more precise than half the sun diameter.
A few years ago I measured my latitude based on the highest angle of the sun. Got it accurate to within half a degree (that is the width of the half-shadow region) with the shadow of a corner of a building, stones, a rope, pen and paper as the only tools (no ruler, no calculator). Observing the location where the sun rises should be much more accurate. Longitude needed a modern clock, of course.
Thanks very much for all the replies.
Has anyone tried timing the Sun or Sirius to get the length of a year?
If anybody can give me some tips, I’ll give it a go myself.
Getting 365.25… Day’s without any modern equipment like an accurate clock and a precision mounted telescope seems impossible, how did the Romans achieve it?
[QUOTE=”Jim60, post: 5447466, member: 590906″]Why did they use the moon as a clock if they knew the year was 365.25 days long, and more importantly, how did they work it out?
It must have been a remarkable achievement for a Babylonian. What was his name?[/QUOTE]
The Babylonians used the moon as a clock because it was important to their religious practices, not because they wanted to be good astronomers, which they were.
Each of the gods worshipped by the Babylonians was associated with a certain month in their calendar, which is quite a bit more complex than modern calendars at reckoning the days.
[URL]https://en.wikipedia.org/wiki/Babylonian_calendar[/URL]
Time over long periods was reckoned using a lunar-solar cycle, which repeats approximately every 19 years.
Certain religious aspects in current times are still based on the moon, like determining the date of Passover or Easter. For centuries, the method used to determine when Passover was celebrated was kept secret by the rabbis, and various methods were developed to calculate the date of Easter either in the Julian or Gregorian calendars. The mathematician C.F. Gauss developed one such algorithm for calculating the date of Easter early in his career.
[URL]https://en.wikipedia.org/wiki/Computus[/URL]
[URL]http://www.staff.science.uu.nl/~gent0113/hovo/downloads/text1_08b.pdf[/URL]
[QUOTE=”Jim60, post: 5447384, member: 590906″]Finding how many days in a year by using the sighting of the crescent moon must have been impossible?
If you count 6 times 29 and 6 times 30, add them together, it comes to 354 days.
That’s 11.25 days less than the accepted figure of 365.25…
Did they know how to add fractions?
It must have been after Newton and Kepler when they finally got it about right?[/QUOTE]
The Greeks knew that the year was just a little shorter than 365.25 days. When Julius Caesar came to power in Rome, he instituted a reform of the calendar, since the old Roman lunar calendar had only about 355 days in a year, and the Roman months had gotten badly out of step with the seasons over time.
To help him reform the Roman calendar to make it more accurate and keep it in step with the seasons, Caesar used the services of a Greek astronomer from Alexandria, one Sosigenes, as explained in this article:
[URL]https://en.wikipedia.org/wiki/Julian_calendar[/URL]
The reformed Julian calendar established the familiar lengths of the months which we still use and created an extra leap day which is inserted into the calendar every four years, the famous Feb. 29. This made the average length of the Julian year some 365.25 days, and the new Julian calendar was a smashing success, at least for a while.
While the length of the actual year is not exactly 365.25 days, the difference amounts to only a few minutes per year, which is almost completely unnoticeable. However, over time, a few minutes here and a few minutes there add up, so that by the middle of the 16th century A.D., the Julian calendar was falling out of step with the seasons, just like the old Roman calendar had, which vexed Caesar. A new set of calendar reforms was established by Pope Gregory XIII in 1582, which eliminated 10 days from the Julian calendar to re-align the months with the seasons, and further adjusted which years received a leap day.
[URL]https://en.wikipedia.org/wiki/Gregorian_calendar[/URL]
Because this Gregorian calendar was a product of the Pope in Rome, it was adopted immediately only in the countries which were Roman catholic. Protestant countries like England and eastern Orthodox countries like Russia continued to use the Old Style Julian calendar to reckon the years. Eventually, England (including the American colonies) adopted the Gregorian calendar in 1752, and Russia held out until 1918, when the Bolsheviks finally decreed calendar reform in Russia.
You can measure the maximal height above the horizon, or the point at the horizon where Sirius goes up/down. You can do the same thing with the sun. Nearly everything that does not depend on the moon or other planets will show the same period of 365.25 days if observed over a few years.
Thanks for the reply.
I read somewhere that the Romans used the star Sirius to calculate the days in a year, but it didn’t give the method they used, or I might have forgotten.
I don’t think I can answer any of those, sorry!
Why did they use the moon as a clock if they knew the year was 365.25 days long, and more importantly, how did they work it out?
It must have been a remarkable achievement for a Babylonian. What was his name?
[QUOTE=”Jim60, post: 5447384, member: 590906″]Finding how many days in a year by using the sighting of the crescent moon must have been impossible?
If you count 6 times 29 and 6 times 30, add them together, it comes to 354 days.
That’s 11.25 days less than the accepted figure of 365.25…
Did they know how to add fractions?
It must have been after Newton and Kepler when they finally got it about right?[/QUOTE]
Yes, ancient civilizations knew how to add fractions, and I doubt they split the year up into 6 months of 29 days and 6 months of 30 days. Being 10+ days off in a single year was very, very noticeable to those who watch for these kinds of things.
[URL]https://en.wikipedia.org/wiki/Fraction_(mathematics)#History[/URL]
Finding how many days in a year by using the sighting of the crescent moon must have been impossible?
If you count 6 times 29 and 6 times 30, add them together, it comes to 354 days.
That’s 11.25 days less than the accepted figure of 365.25…
Did they know how to add fractions?
It must have been after Newton and Kepler when they finally got it about right?
It is quite easy to get the length of a year with a precision of a day over a human lifetime without any dedicated measurements, and getting 365.25 as approximation is possible with very simple astronomical observations. I don’t think anyone ever thought a year would be 360 days long. The Babylonians certainly knew it better (see above).
Getting more precise than 365.25 is challenging, and brings up all the complications mentioned in the insights article.
[QUOTE=”Jim60, post: 5446401, member: 590906″]In the past they must have thought there was only 360 days in a year, that’s probably why we’ve got 360 degrees in a circle?[/QUOTE]
Nah, the 360 degrees thing is left over from the Babylonian number system: [URL]https://en.wikipedia.org/wiki/Babylonian_numerals[/URL]
[I]
The legacy of sexagesimal still survives to this day, in the form of [URL=’https://en.wikipedia.org/wiki/Degree_(angle)’]degrees[/URL] (360° in a [URL=’https://en.wikipedia.org/wiki/Circle’]circle[/URL] or 60° in an [URL=’https://en.wikipedia.org/wiki/Angle’]angle[/URL] of an [URL=’https://en.wikipedia.org/wiki/Equilateral_triangle’]equilateral triangle[/URL]), [URL=’https://en.wikipedia.org/wiki/Minute’]minutes[/URL], and [URL=’https://en.wikipedia.org/wiki/Second’]seconds[/URL] in [URL=’https://en.wikipedia.org/wiki/Trigonometry’]trigonometry[/URL] and the measurement of [URL=’https://en.wikipedia.org/wiki/Time’]time[/URL], although both of these systems are actually mixed radix.[/I]
The Babylonian calendar used lunar months, which vary from 29-30 days each, along with an intercalary month when necessary. The years wouldn’t have been 360 days long as far as I can tell.
In the past they must have thought there was only 360 days in a year, that’s probably why we’ve got 360 degrees in a circle?
Excellent post. I’d like to point out that we live in a very special time during the 26,000 precession cycle where we have the brightest star that it’s possible to have as the North Star which is very close to the actual north pole, this lasts about 200 years before drifting away.
Well, try this. Figure out the EXACT way that they determine when Easter is each year. And, no, it has little to do with the Vernal Equinox. It has something to do with “The Golden Number” (which has a lot to do with astronomy). This was so confusing that the various churches (Catholic, Anglican, etc.) would print the day Easter falls on for a large number of years in the Common Prayer Book of the ordinary Christian (Easter is by far the most important Christian holiday).
Three historical definitions of “day” are given, but the currently excepted definition is not included i.e.
86,400 SI seconds where the SI second is based on the hyperfine transition frequency of the caesium atom.
Really an interesting topic though I never thought it would have been so much elaborated. I know only the direct 365 days :p
[QUOTE=”mfb, post: 5118878, member: 405866″]You need constants with dimensions to fix units. Dimensionless constants (in SI) do not allow to define things like kilograms.[/QUOTE]
Ahh, right, silly me. But incorporation of the fundamental constants in SI definitions is definitely the need of the hour.
[QUOTE=”PWiz, post: 5118877, member: 536763″]Well thank god they realized the slowing rotation rate of the Earth in the 60’s and changed the SI definition a second. :rolleyes:[/quote]Well atomic clocks would run more stable than Earth either way. But frequently tuning them would be really messy and make astronomic measurements weird (“and then we have to account for the longer seconds in 1994 to compare the results”).
[quote]I guess we just have to wait for the Kilogram, Ampere and Kelvin to get a much needed change in definition to one in terms of (preferably dimensionless) fundamental constants.[/QUOTE]You need constants with dimensions to fix units. Dimensionless constants (in SI) do not allow to define things like kilograms.
[QUOTE=”mfb, post: 5118871, member: 405866″]To make things worse, the rotation rate of earth is not constant. The effect can be up to one second per year.
Over longer timescales, the days get longer by about 2.3 milliseconds each century, currently with a lower rate of 1.7 ms as some continental masses still move around related to the last ice age.
To keep atomic clocks in sync with the position of the sun, we frequently need leap seconds, and we will need more and more in the future.[/QUOTE]
Well thank god they realized the slowing rotation rate of the Earth in the 60’s and changed the SI definition a second. :rolleyes:
I guess we just have to wait for the Kilogram, Ampere and Kelvin to get a much needed change in definition to one in terms of (preferably dimensionless) fundamental constants.
To make things worse, the rotation rate of earth is not constant. The effect can be up to one second per year.
Over longer timescales, the days get longer by about 2.3 milliseconds each century, currently with a lower rate of 1.7 ms as some continental masses still move around related to the last ice age.
To keep atomic clocks in sync with the position of the sun, we frequently need leap seconds, and we will need more and more in the future.
I just came across this. There is a very minor error in the following:"you’d eventually have Summer weather occurring in December. The end result is that the tropical year is slightly shorter than the sidereal year, being 365.24219 mean solar days long vs. 365.2563662 mean solar days to a sidereal year. 365.2563662 times 400 equals 146096.876, which is very close to the 146097 days in the Gregorian calendar for the same period, which is why it is such a better fit."The logic (and reality) suggest you meant to say: 365.24219 times 400 equals 146096.876
Excellent post. I'd like to point out that we live in a very special time during the 26,000 precession cycle where we have the brightest star that it's possible to have as the North Star which is very close to the actual north pole, this lasts about 200 years before drifting away.
Three historical definitions of "day" are given, but the currently excepted definition is not included i.e.86,400 SI seconds where the SI second is based on the hyperfine transition frequency of the caesium atom.
"This also means that the position of the Sun with respect to the stars on a given date as seen from the Earth also changes over the years. In the time since the dates for the astrological signs were established, the Sun has moved an entire Zodiac sign. Thus, during the Summer solstice this year, instead of just entering Cancer as the astrological dates suggest, the Sun is just leaving Taurus and entering Gemini. (So if the the Tropic of Cancer had been named today, it would have likely been called the Tropic of Gemini and the Tropic of Capricorn would be the Tropic of Sagittarius)."So, at the time astrological signs were established, the vernal equinox pointed towards the constellation Aries. And that some point in Aries (the first point of Aries?) could be used to find the direction of the vernal equinox no matter what day of the year it was. And now the direction of the vernal equinox (the first point of Aries) lies in the constellation Pisces?Man, I can't wait until the direction of the vernal equinox (the first point of Aries) approaches the constellation Aquarius. It will be such a great occasion that we'll all dance and sing songs about it! "It is the dawning of the Age of Aquarius!"
Great post, enjoyable read. Thank you.
This is very interesting topic.I never knew that and never thought about that.
[quote]This means that the Earth must continue rotating a bit longer (~ 4 min) to finally return that spot to directly under the Sun. So it turns out that 1 solar day equals 1.002737803 stellar days.[/quote]Okay.That means a solar day is longer than a stellar day, or equivalently a stellar day is shorter than a solar day. A solar day is 86400 seconds. How can a stellar day be 86164 seconds then?
This is a very cool topic. I recently read a book by Stephen Jay Gould called "Millenium" (or something like that) which discussed many of these points. The calendar is a formidably complicated beast to tackle when one really gets deep into the topic.
There are so many rational ways to measure a year. But let's not forget our Muslim readers. Their year is based on how many months have passed. And their months are based on direct observation of the moon. The length of the month can change depending on the weather since clouds interfere with moon sightings. Because of this it is impossible to predict with certainty what the next Muslim year's dates will be. The future is spun out real time.So anyone who thinks these various astronomical measurements are complex, be glad you're not a Muslim. (Unless you are, in which case the future is as God wills it.)Humanity has measured time in lots of ways in history. Some of them will seem strange to us living in an era where calendars are common place technology.
Great first entry @janus! This is a topic I never knew could be so deep!