earth year

How to Scientifically Measure How Many Days Are in a Year

Estimated Read Time: 11 minute(s)
Common Topics: year, sun, days, sidereal, day

Days in a Year 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 that a year has 365 days, with leap years having 366 days.
  • The concept of a “stellar day,” based on Earth’s rotation concerning fixed stars, is introduced.
  • The “sidereal year,” the time it takes for Earth to orbit the Sun concerning the stars, is discussed, revealing a slightly longer duration than previously thought.
  • Differences between stellar and 24-hour days are attributed to the Earth’s elliptical orbit and its effect on solar day length.
  • Precession, the wobbling of the Earth’s axis, is explained as 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?

earth
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. 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, concerning what? The reference often used by astronomers is that of the “fixed” stars. If we measure one rotation of the Earth concerning the stars we call it a “stellar day”. Using the same reference, one trip of the Earth around the Sun concerning 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 concerning the stars would be called a sidereal day, but oddly, a sidereal day is measured concerning a different reference which we will deal with later on.)

If you measure the Earth’s rotation concerning the stars you get 86164.098903691 seconds per rotation, and if you do the same for the Earth’s orbit around the Sun, you get 31558149.76… sec. This means that, concerning 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 24-hour clock is based on the “Solar day”, or the time it takes for the Sun to go from noon on one day to noon on the next, and this period is affected by the Earth’s orbit around the Sun.

The Stellar vs. Solar Day

In the time it takes the Earth to make one rotation concerning 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 to 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

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 is a result of 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.

Gregorian year vs 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 concerning 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 the 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 noticeable from one year to the next, but throughout 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 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.)

Answering “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

The Shifting Perihelion

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.


75 replies
  1. Jim60 says:

    Think I’ve figured out how Wikipedia got 365.256363004 SI days for the Sidereal year?
    Earths semi-major axis a = 1.4959974465490E+11 Metres.
    Gravitational constant G = 6.67408E-11 Newton Metres squared / kilogram square
    Suns mass M1 = 1.98855E+30kg
    Earths mass M2 = 5.97237E+24kg
    T = 2*PI()*SQRT(a^3/G*(M1+M2))
    Orbital period T = 31558149.76354540 SI seconds

  2. Jim60 says:

    2000 to 2025 aphelion to aphelion
    2000 July 3rd 23 hours 49 minutes – 2001 July 4th 13 hours 37 minutes 365.57500 SI Days
    2001 July 4th 13 hours 37 minutes – 2002 July 6th 3 hours 47 minutes 366.590278 SI days
    2002 July 6th 3 hours 47 minutes – 2003 July 4th 5 hours 40 minutes 363.078472 SI Days
    2003 July 4th 5 hours 40 minutes – 2004 July 5th 10 hours 54 minutes 367.218056 SI Days
    2004 July 5th 10 hours 54 minutes – 2005 July 5th 4 hours 58 minutes 364.752778 SI Days
    2005 July 5th 4 hours 58 minutes – 2006 July 3rd 23 hours 10 minutes 363.758333 SI Days
    2006 July 3rd 23 hours 10 minutes – 2007 July 6th 23 hours 53 minutes 368.029861 SI Days
    2007 July 6th 23 hours 53 minutes – 2008 July 4th 7 hours 41 minutes 363.325 SI Days
    2008 July 4th 7 hours 41 minutes – 2009 July 4th 1 hour 40 minutes 364.749306 SI Days
    2009 July 4th 1 hour 40 minutes – 2010 July 6th 11 hours 30 minutes 367.409722 SI days
    2010 July 6th 11 hours 30 minutes – 2011 July 4th 14 hours 54 minutes 363.141667 SI days
    2011 July 4th 14 hours 54 minutes- 2012 July 5th 3 hours 32 minutes 366.526389 SI days
    2012 July 5th 3 hours 32 minutes – 2013 July 5th 14 hours 44 minutes 365.466667 SI days
    2013 July 5th 14 hours 44 minutes – 2014 July 4th 0 hours 13 minutes 363.395139 SI days
    2014 July 4th 0 hours 13 minutes – 2015 July 6th 19 hours 40 minutes 367.810417 SI days
    2015 July 6th 19 hours 40 minutes – 2016 July 4th 16 hours 24 minutes 363.863889 SI days
    2016 July 4th 16 hours 24 minutes – 2017 July 3rd 20 hours 11 minutes 364.157639 SI days
    2017 July 3rd 20 hours 11 minutes – 2018 July 6th 16 hours 47 minutes 367.858333 SI day
    2018 July 6th 16 hours 47 minutes – 2019 July 4th 22 hours 11 minutes 363.225 SI day
    2019 July 4th 22 hours 11 minutes – 2020 July 4th 11 hours 35 minutes 365.558333 SI days
    2020 July 4th 11 hours 35 minutes – 2021 July 5th 22 hours 27 minutes 366.452778 SI Days
    2021 July 5th 22 hours 27 minutes – 2022 July 4th 7 hours 11 minutes 363.363889 SI days
    2022 July 4th 7 hours 11 minutes – 2023 July 6th 20 hours 7 minutes 367.538889 SI Days
    2023 July 6th 20 hours 7 minutes – 2024 July 5th 5 hours 6 minutes 364.374306 SI Days
    2024 July 5th 5 hours 6 minutes – 2025 July 3rd 19 hours 55 minutes 363.617361 SI Days
    Total 9130.8375 SI days
    2000 to 2025 Average year 365.2335 SI Days
    Still nowhere near Wikipedia Anomalistic year of 365.259363 days.
    Does anyone know how they calculated this year?

  3. Jim60 says:

    “I merged your posts, you can edit your posts if you want to add something.There are databases keeping track of those dates.”

    Do you have a link?

  4. mfb says:

    I merged your posts, you can edit your posts if you want to add something.”What way could that be proved?”There are databases keeping track of those dates.

  5. Jim60 says:

    It depends on how accurate the data is.
    According to that site I’ve been using, Aphelion, Earths closes approach to the Sun will occur in 62.612 SI days on July 4th 16 hours 24 minutes UT.
    What way could that be proved?

    Sorry I meant, Earth will be furthest away from the Sun.

    Note to self, must engage brain before posting.

    And finally I hope I’ve got it right, its 61.42 SI days away on July the 4th 16 hours 24 minutes.

  6. Jim60 says:

    I’m trying to find the average anomalistic year and the average tropical year.
    Maybe I need more than 25 years?
    Does anyone know of a data source that I can use?

  7. Jim60 says:

    Wikipedia average Anomalistic year is 365.259636 days, but using that site again.
    Perihelion Perihelion Anomalistic year
    2000 Jan 3rd 5 hours 18 minutes- 2001 Jan 4th 8 hours 52 minutes 367.1486111 days
    2001 Jan 4th 8 hours 52 minutes – 2002 Jan 2nd 14 hours 9 minutes 363.2201389 days
    2002 Jan 2nd 14 hours 9 minutes- 2003 Jan 4th 5 hours 2 minutes 366.6201389 days
    2003 Jan 4th 5 hours 2 minutes – 2004 Jan 4th 17 hours 42 minutes 365.5277778 days
    2004 Jan 4th 17 hours 42 minutes – 2005 Jan 2nd 0 hours 35 minutes 363.2868056 days
    2005 Jan 2nd 0 hours 35 minutes – 2006 Jan 4th 15 hours 30 minutes 367.6215278 days
    2006 Jan 4th 15 hours 30 minutes – 2007 Jan 3rd 19 hours 43 minutes 364.1756944 days
    2007 Jan 3rd 19 hours 43 minutes – 2008 Jan 2nd 23 hours51 minutes 364.1722222 days
    2008 Jan 2nd 23 hours 51 minutes – 2009 Jan 4th 15 hours 3 minutes 367.6520833 days
    2009 Jan 4th 15 hours 30 minutes – 2010 Jan 3rd 0 hours 9 minutes 363.3604167 days
    2010 Jan 3rd 0 hours 9 minutes – 2011 Jan 3rd 18 hours 32 minutes 365.7659722 days
    2011 Jan 3rd 18 hours 32 minutes – 2012 Jan 5th 0 hours 32 minutes 366.25 days
    2012 Jan 5th 0 hours 32 minutes – 2013 Jan 2nd 4 hours 38 minutes 363.1708333 days
    2013 Jan 2nd 4 hours 38 minutes – 2014 Jan 4th 11 hours 59 minutes 367.30625 days
    2014 Jan 4th 11 hours 59 minutes – 2015 Jan 4th 6 hours 36 minutes 364.7756944 days
    2015 Jan 4th 6 hours 36 minutes – 2016 Jan 2nd 22 hours 49 minutes 363.6756944 days
    2016 Jan 2nd 22 hours 49 minutes – 2017 Jan 4th 14 hours 18 minutes 367.6451389 days
    2017 Jan 4th 14 hours 18 minutes – 2018 Jan 3rd 5 hours 35 minutes 363.6368056 days
    2018 Jan 3rd 5 hours 35 minutes – 2019 Jan 3rd 5 hours 20 minutes 364.9895833 days
    2019 Jan 3rd 5 hours 20 minutes – 2020 Jan 5th 7 hours 48 minutes 367.1027778 days
    2020 Jan 5th 7 hours 48 minutes – 2021 Jan 2nd 13 hours 51 minutes 363.2520833 days
    2021 Jan 2nd 13 hours 51 minutes – 2022 Jan 4th 6 hours 55 minutes 366.7111111 days
    2022 Jan 4th 6 hours 55 minutes – 2023 Jan 4th 16 hours 17 minutes 365.3902778 days
    2023 Jan 4th 16 hours 17 minutes – 2024 Jan 3rd 0 hours 39 minutes 363.3486111 days
    2024 Jan 3rd 0 hour 39 minutes – 2025 Jan 4th 13 hours 28 Minutes 367.5340278 days

    2000 to 2025 average Anomalistic year 365.3336111110920 days

  8. Jim60 says:

    Using that site again to count the days from 2000 December 21st 13 hours 37 minutes to 2025 December 21st 15 hours 3 minutes (winter solstice to winter solstice),
    Totalled 9131.05722224 SI days, divided by 25 years is 365.24239 SI days per year, minus the average tropical year of 365.24219 days from this tropical year of 365.24239 days is 17.18 seconds more.

  9. Jim60 says:

    Using that site again to count the days from 2000 June 21st 1 hour 48 minutes to 2025 June 21st 2 hours 42 minutes (summer solstice to summer solstice).

    Totalled 9131.0375 SI days, divided by 25 years is 365.24150 SI days per year, minus this value from the average Tropical year of 365.24219 days is 59.616 seconds less.

  10. PAllen says:

    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

  11. Jim60 says:

    Using that site again to find the average Tropical year from 2000 March 20th 7 hours 35 minutes to 2025 March 20th 9 hours 1 minute, I got 9131.0597222224 SI days.

    Dividing by 25 years; the average tropical year from 2000 to 2025 is 365.24238888890 SI days.

    Wikipedia value for the average Tropical year is 365.24219 days, and as you stated in your post DH, that is very close to the average tropical year from 2000 to 2025.

    I get it to just over half a second averaged over 25 years

  12. Jim60 says:

    Looking at that site and taking the Perihelion to Perihelion dates from Jan the 3rd 5 hours 18 minutes 2000 to Jan 4th 13 hours 28 minutes 2025.

    Using your technique DH, I get a difference in time of 9133.3402777771 SI days, when I divided the total days by 25 years I get a value of 365.3336111 days.

    Minus the Anomalistic year that I got from Wikipedia of 365.259363 days from 365.3336111, I get a difference of 106.52416 minutes over 25 years.

    I then divided these minutes by 25 and got a value of 4.260966398 minutes per year.

    Is this the anomalistic year happening later in time by this value per year, or have I got it totally wrong?

  13. Old Tele man says:

    “The easy part: An SI day is 86400 SI seconds. The word “mean” in “mean solar day” means “average.” A mean solar day is the amount of time from one solar noon to the next, averaged over the course of a year. The SI second and the 86400 second long day was based on the concept of a mean solar day. (A more technical aspect involves something called the “fictitious mean Sun”. I won’t go into that.)

    I wrote “was” because that definition is not what’s used anymore. Scientists have noticed for quite some time that even after accounting for the equation of time (google that term), the length of a day (one solar noon to the next) is not constant. The definition of the second was based on data collected over the course of about 150 years centered on 1820. 196 years later (i.e., now), the average time from one solar noon to the next is about 86400.002 seconds, or 2 milliseconds longer than the 86400 seconds it took in 1820. The Earth’s rotation rate is very gradually slowing down.”
    …a side-effect of our fleeting moon’s slowly increasing orbital distance.

  14. D H says:

    “What’s the difference in time from a mean solar day and a SI day? It’s confusing.”
    The easy part: An SI day is 86400 SI seconds. The word “mean” in “mean solar day” means “average.” A mean solar day is the amount of time from one solar noon to the next, averaged over the course of a year. The SI second and the 86400 second long day was based on the concept of a mean solar day. (A more technical aspect involves something called the “fictitious mean Sun”. I won’t go into that.)

    I wrote “was” because that definition is not what’s used anymore. Scientists have noticed for quite some time that even after accounting for the equation of time (google that term), the length of a day (one solar noon to the next) is not constant. The definition of the second was based on data collected over the course of about 150 years centered on 1820. 196 years later (i.e., now), the average time from one solar noon to the next is about 86400.002 seconds, or 2 milliseconds longer than the 86400 seconds it took in 1820. The Earth’s rotation rate is very gradually slowing down.

  15. D H says:

    “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.”
    Actually, it’s about 10 minutes more than it “should” be. A tropical year is about 365 days, 5 hours, 48 minutes, and 45 seconds long, or 365.24219 days.

    “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.”
    I used “should” in quotes because there is something else that needs to be taken into account. That something else is the Moon. The Moon’s orbit about the Earth is slightly inclined compared to the Earth’s orbit about the Sun. Suppose that on one March 20th the Moon is above the ecliptic but below on the next (or vice versa). This changes the timing of the equinoxes from year to year by several minutes.

    One has to average over a longer span of time to see something close to 365.24219 days from vernal equinox to vernal equinox. For example, the link you used lists equinoxes and solstices from 2000 to 2025. The time from the 2000 vernal equinox to the 2025 vernal equinox is 25*365+6 days, 1 hour, and 26 minutes. Divide that by 25 and you get 365.2424 days, which is quite close to the long term average of 365.24219 days.

  16. Jim60 says:

    “0.01 year is more than 3.5 days.

    The tropical year is about 365.242 mean solar days, as of 2010.
    The sidereal year is about 365.256 SI days, for the J2000 epoch.

    The difference is about 0.01 day, not 0.01 year.

    [URL]https://en.wikipedia.org/wiki/Tropical_year[/URL]
    [URL]https://en.wikipedia.org/wiki/Sidereal_year[/URL]”

    What’s the difference in time from a mean solar day and a SI day? It’s confusing.

  17. SteamKing says:

    “Tropical year, but the difference to the sidereal year is only 0.01 years, which needs the mentioned equatorial ring or similar devices to detect, otherwise the difference does not matter.”
    0.01 year is more than 3.5 days.

    The tropical year is about 365.242 mean solar days, as of 2010.
    The sidereal year is about 365.256 SI days, for the J2000 epoch.

    The difference is about 0.01 day, not 0.01 year.

    [URL]https://en.wikipedia.org/wiki/Tropical_year[/URL]
    [URL]https://en.wikipedia.org/wiki/Sidereal_year[/URL]

  18. mfb says:

    Tropical year, but the difference to the sidereal year is only 0.01 [s]years[/s] days, which needs the mentioned equatorial ring or similar devices to detect, otherwise the difference does not matter.

  19. mfb says:

    “You make sound easy?”It [b]is[/b] easy.
    Manhattan has great reference points and good horizon sight with the street layout, [url=https://en.wikipedia.org/wiki/Manhattanhenge]the effect is quite famous there[/url]. Note how “full sun” has a deviation of at most one day. The same thing can be done with any other reference point close to the horizon.

    Getting 365 days with a bit of care is easy, and getting 365.25 with a longer measurement time is not hard either.

  20. jim mcnamara says:

    Time lapse exposures of the analemma:

    [URL]https://www.google.com/imgres?imgurl=https://i.ytimg.com/vi/jQT5XRdrqvw/maxresdefault.jpg&imgrefurl=https://www.youtube.com/watch%3Fv%3DjQT5XRdrqvw&h=3000&w=2000&tbnid=u7lbbU0x_FZQsM:&tbnh=160&tbnw=106&docid=kfJVkk2dx49PcM&itg=1&usg=__MRSlpWmrpu6-o-YNI8q3gi-2tJg=&sa=X&ved=0ahUKEwi62ISM56fMAhUL9WMKHeG4CGIQ9QEIIzAA[/URL]

  21. Vanadium 50 says:

    “Has anyone on this forum attempted to count the days in a year using this method, or any other way for that matter,”

    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.

  22. Jim60 says:

    “Only if you want to observe the sunrise every day in the year, which is not necessary.”
    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?

  23. mfb says:

    “That would mean you would need a total angle of about 90 degrees clear view of the horizon.”Only if you want to observe the sunrise every day in the year, which is not necessary.

  24. Jim60 says:

    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.

  25. anorlunda says:

    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.

  26. Jim60 says:

    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?

  27. mfb says:

    It should not be 365.25.

    Independent of that, there are smaller variations, some regular, some not.”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?”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

  28. Jim60 says:

    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.

  29. Jim60 says:

    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.

  30. Drakkith says:

    “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?”

    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.

  31. Jim60 says:

    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?

  32. Jim60 says:

    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.

  33. mfb says:

    “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?”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.

  34. SteamKing says:

    “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?”
    As far as I know, the sun rises in the east and sets in the west every 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?”

    Hipparchus and other Greek astronomers used a device called an equatorial ring to determine when the equinoxes occurred:

    [URL]https://en.wikipedia.org/wiki/Equatorial_ring[/URL]

  35. Jim60 says:

    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?

  36. SteamKing says:

    “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?”
    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.

  37. mfb says:

    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.

  38. Jim60 says:

    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?

  39. SteamKing says:

    “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?”
    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]

  40. SteamKing says:

    “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?”
    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.

  41. mfb says:

    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.

  42. Jim60 says:

    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.

  43. Jim60 says:

    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?

  44. Drakkith says:

    “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?”

    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]

  45. Jim60 says:

    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?

  46. mfb says:

    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.

  47. Drakkith says:

    “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?”

    Nah, the 360 degrees thing is left over from the Babylonian number system: [URL]https://en.wikipedia.org/wiki/Babylonian_numerals[/URL]

    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.

    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.

  48. bland says:

    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.

  49. Edward Oxford says:

    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).

  50. the_emi_guy says:

    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.

  51. PWiz says:

    “You need constants with dimensions to fix units. Dimensionless constants (in SI) do not allow to define things like kilograms.”
    Ahh, right, silly me. But incorporation of the fundamental constants in SI definitions is definitely the need of the hour.

  52. mfb says:

    “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.”You need constants with dimensions to fix units. Dimensionless constants (in SI) do not allow to define things like kilograms.

  53. PWiz says:

    “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.”
    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.

  54. mfb says:

    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.

  55. PAllen says:

    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

  56. bland says:

    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.

  57. the_emi_guy says:

    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.

  58. BobG says:

    "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!"

  59. mfb says:

    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.

    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?

  60. QuantumCurt says:

    This is a very cool topic. I recently read a book by Stephen Jay Gould called “Millennium” (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.

  61. Jeff Rosenbury says:

    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 in 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 commonplace technology.

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