<blockquote>
<p>if the sun appears to be in the exact point in the sky as that time the previous solar year,</p>
</blockquote>
<p>This statement means the sun is at (nearly) the same position <em>relative to the stars</em> that it was the previous year.  Not that the sun is at the same point in the sky relative to the earth's surface.</p>
<p>So there are two motions: the sun across the field of stars, and the earth's direction to see things.</p>
<p>For the sidereal year (which is <em>really</em> close to the solar year), all we care about is where the sun has moved relative to the stars.  The angle of earth's rotation so that your house is able to see the sun doesn't matter.</p>

CC BY-SA 4.0
2023-04-10T17:59:36.743Z
2023-04-10T17:59:36.743Z
BowlOfRed
<p>Let‘s turn your question around: Why should they be in sync?</p>
<p>The only mechanism that redistributes angular momentum between two celestial bodies are tidal forces (not considering general relativity and crashing the bodies against each other. Angular momentum of the whole system must be conserved). In the long term this leads to locked rotation as in the cases of moon, Mercury and between Pluto and Charon. As you probably know, tides created by sun and moon are creating a <a href="https://en.wikipedia.org/wiki/Tidal_acceleration" rel="noreferrer">drag on earth to go slower</a> and earth’s rotation was quicker in the past. Note that this force in the end is going to force earth into a locked rotation around sun (cf. spring tide). So we will end up with a day length of one year (if the force is strong enough compared to other influences on angular momentum like meteors entering earth atmosphere).</p>
<p>Simply put: There‘s a mechanism to create a resonance between day length and year length, but the day length is going to become 1 year. That it doesn't match an integer number of days at the current point in time is coincidence.</p>
<p>Note that <a href="https://en.wikipedia.org/wiki/Orbital_resonance" rel="noreferrer">stronger resonances</a> do exist between planet orbits (e.g. Jupiter‘s moons), which is like pushing gravitationally a child on a swing, but these forces do not couple to the angular momentum of the respective body and the tidal forces are much smaller in comparison.</p>

CC BY-SA 4.0
2023-04-11T11:19:58.730Z
*There‘s no mechanism to create a resonance between day length and year length*, except that your previous paragraph explained that there is, it's just too weak between Sun and Earth to lead to resonance within the lifetime of the solar system.
gerrit
CC BY-SA 4.0
2023-04-11T15:08:11.960Z
Agree this answer seems to veer around in what the takeaway should be. It goes from "what mechanism could possibly be responsible" to "here is a mechanism that could do it and examples where it happens" and then concludes with "there is no mechanism, rotational and orbital period are unrelated", which rather contradicts everything else.
Nuclear Hoagie
CC BY-SA 4.0
2023-04-12T12:48:33.857Z
@gerrit thanks for catching my lapse. Hope I fixed it now.
Grimaldi
CC BY-SA 4.0
2023-04-10T16:14:31.543Z
2023-04-12T12:48:12.600Z
2023-04-12T12:48:12.600Z
Grimaldi
<p>It isn't in the same place.</p>
<p>We can measure &quot;one year&quot; starting at any point in time, suppose we start at sunrise on January 1st at some convenient location (eg London). We know where the sun is; it is on the horizon.  The time is about 8am.</p>
<p>One solar year later (there is a slight detail about the exact meaning of this which I shall gloss over) it is 1:48pm on January 1st, and the sun is not in the same place. Unless there has been a leap day, in which case it is about 1:48 on December 31st.</p>
<p>So you are totally right, after one solar year it is an entirely different time. We set our calendar to start each day at midnight and this means that if we didn't have leap days the calendar would quickly get out of sync with the seasons.</p>
<p>So why not <em>define</em> the solar year to be the time for the sun to reach the same place in the sky?  Let's look at this in more detail.  On Jan 1st 2022 at sunrise (in my location) it is 8:03:25 and the sun has an azimuth of 127.09 degrees (and is on the horizon).  When is it in exactly the same position? Well not on Jan 1st 2023 (it has an azimuth of 127.13), nor on Jan 2nd (126.99) nor Dec 31st (127.28)</p>
<p>The sun doesn't come back to the same place after an integer number of days, it's pretty close <em>but not exact</em>  In fact, what you would find is that after 4 years of 365 days each, the sun would be close to the 127.09 degrees on Jan 2nd.  If you kept on going, after about 700 years, the sun would be rising at that azimuth in the middle of June!</p>
<p>It's just not possible to choose an integer number of days between which the sun returns to <em>exactly</em> the same position.</p>
<p>So the definition of the solar year is the time for the sun to get back to the same position relative to the equinox (the point in the sky where the path of the sun crosses the equator)  This is the &quot;tropical year&quot; it doesn't have an integer number of days, but it does keep the seasons exactly aligned.  If it is the start of spring, then 1000 &quot;tropical years&quot; later it will still be the start of spring (though a different time of day)  And we then design a calendar to approximately align a variable number of days to the tropical year. This is convenient, it means (for example) farmers don't need to adjust for the shift in the calendar when planning the best time for planting.</p>
<p>This is not the only way of designing a calendar.  Most notably the Islamic calendar strictly follows a cycle of twelve lunar months, and so a month such as Ramaḍān may occur in the northern hemisphere summer or winter.</p>

CC BY-SA 4.0
2023-04-11T08:47:38.840Z
It seems to me that this is the only answer that a) understands the question and b) is correct.
Mike Nakis
CC BY-SA 4.0
2023-04-11T09:17:01.223Z
Or possibly we have both misunderstood the question in the same way.
James K
CC BY-SA 4.0
2023-04-11T16:22:29.563Z
@MikeNakis but JamesK **does not** answer the question "Why are there not a whole number of solar days in a solar year?".  He just says "yes" to "Wouldn't that difference of almost 6hrs or 18hrs mean it would be an entirely different time of day than the previous year?"
RonJohn
CC BY-SA 4.0
2023-04-10T05:56:14.540Z
2023-04-20T17:48:22.610Z
2023-04-20T17:48:22.610Z
James K
<p>Take a bike and a race track. Mark one spot on the side of one of your tires.</p>
<p>Now place the bike on the starting line, making sure the mark is precisely at the lowest point of the wheel (where it touches the ground), precisely on the line.</p>
<p>Do one lap, then stop on the starting line.</p>
<p>The chances the mark on your tire will be aligned with the line at that point are quite small (it will happen only if the path you have taken around the track is an exact multiple of the circumference of your wheel).</p>
<p>(aside: on a race track the exact distance around the track will depend on where on the track you ride, so you can get very different results from one lap to the next)</p>
<p>The same happens for the Earth around the Sun. It spins, and a given meridian on the surface of the Earth will point towards the Sun every day (like the mark on the tire will be touching the ground each time the wheel makes one turn), and at the same time, it orbits around the Sun, like your bike doing a lap around the track. No reason for the same meridian to be towards the Sun exactly one year later.</p>
<p>You can see that either as &quot;365 days later, when the reference meridian points towards the Sun, we haven't made exactly one lap around the Sun (not back in the same place on the orbit)&quot; or &quot;exactly a tropical year later, our reference meridian does not point towards the sun&quot;, which really are the same thing (the spot on your tire touches the ground somewhere other than the start line, or when you cross the start line, the spot on your tire is not on the ground).</p>
<p>This answers your specific question:</p>
<blockquote>
<p>However, if the sun appears to be in the exact point in the sky as that time the previous solar year, how does the earth reach this exact position while only being 5hrs, 48 min, 45 secs into its final rotation of the year? Wouldn't that difference of almost 6hrs or 18hrs mean it would be an entirely different time of day than the previous year?</p>
</blockquote>
<p>At a given location, the Sun will <strong>not</strong> appear at the same exact point in the sky as that time the previous solar year. Either it appears at the same exact point in the sky as that time the previous <strong>year</strong> (i.e. 365 days later in most cases), but the Earth hasn't gone through a full orbit around the sun (we haven't crossed the start line on the racetrack yet), or if you are exactly one solar year later (we crossed the line), either it will be a the same place at a different time, or it will be in a different place at the same time.</p>
<p>So after a year we're indeed off by about 6 hours, the next year about 12, then third one, 18, and four years later, about 24 hours. So we are off by a day. Would seem we're back on our feet (our reference meridian again points towards the sun when we reach that famous spot on the orbit). But if we didn't do anything about that, after a (long) while, winter would happen in May.</p>
<p>So <a href="https://en.wikipedia.org/wiki/Julian_calendar" rel="noreferrer">some people determined a couple of millennia ago</a> that we would need to make a few adjustments now and then to the duration of a year, and introduced the leap years. So as we are off by about a day after 4 years, every 4 years the year is one day longer, and we're back on our feet.</p>
<p>Of course <a href="https://en.wikipedia.org/wiki/Gregorian_calendar" rel="noreferrer">some other people noticed a few centuries later</a> that this was not exact (it's not precisely 365.25 days), and more adjustments were made (so we skip leap years on years that are multiples of 100, but not multiples of 400). Still not exact either, but a lot closer.</p>
<p>Guess what, the minor difference between those two actually resulted in a difference of 13 days between the two calendars, and that introduced quite a bit of confusion for quite a while as not everybody switched calendars at the same time.</p>
<p>As a parallel, it may be interesting to note that some cultures use calendars based on the Moon rather than the Sun. Here again, no reason for the orbit of the Earth around the Sun (a year) to be a multiple of the orbit of the Moon around the Earth (about 29.5 days -- again, not a multiple of the time it takes the Earth to rotate around its axis either). This has the consequence that such calendars either need to add some extra days to &quot;re-align&quot; on the year, or they just drift year after year (which explains why the month of Ramadan in the Islamic Calendar occurs in March/April this year, while back in 2011 it started in August).</p>

CC BY-SA 4.0
2023-04-11T16:54:21.067Z
2023-04-11T16:54:21.067Z
jcaron
<p>The motion of the sun in the sky doesn't exactly repeat at one-year intervals, but it's still reasonable to say that it's on a one-year cycle.</p>
<p>If you record the sun's position at the same time each day, the points will appear to lie on a figure-8 curve (<a href="https://en.wikipedia.org/wiki/Analemma" rel="noreferrer">analemma</a>). If you parametrize the continuous curve by a real number that runs from 0 to 365.25, and your first measurement is at 0, then the points you plot will be 0, 1, 2, ..., 364, 365, 0.75, 1.75, ..., 363.75, 364.75, 0.5, 1.5, .... This repeats after four years, but of course if I'd used a more accurate estimate of the year length it wouldn't. More importantly, it <em>doesn't</em> repeat after <em>one</em> year. But it is clearly cyclic with a period of one year, nevertheless.</p>

CC BY-SA 4.0
2023-04-10T04:56:12.000Z
2023-04-10T04:56:12.000Z
benrg
<p>We do have one planet in our solar system with a neat <a href="https://en.wikipedia.org/wiki/Orbital_resonance" rel="noreferrer">orbital resonance</a>: <a href="https://en.wikipedia.org/wiki/Mercury_(planet)" rel="noreferrer">Mercury</a>, with an orbital period of 87.9691 Earth days, and a sidereal rotation period of 58.646 days, which works out to almost exactly 1.5 sideral days, or 0.5 solar day, in its year.  Being the closest planet to the Sun,  it experiences a stronger gravitational pull than the other planets, thus the most affected by <a href="https://en.wikipedia.org/wiki/Tidal_locking" rel="noreferrer">tidal locking</a>.</p>
<p>But Earth is farther away, and has its rotation affected by competing factors that Mercury doesn't have: a large Moon with its own gravitational pull, atmospheric pressure, and ocean currents.  The length of Earth's day <a href="https://en.wikipedia.org/wiki/Day_length_fluctuations" rel="noreferrer">isn't even constant</a>, and the rotation is gradually slowing down, so even if there ever were a whole number of days in a year, that wouldn't remain true over the course of millennia.</p>

CC BY-SA 4.0
2023-04-10T23:27:42.090Z
2023-04-10T23:27:42.090Z
dan04
<h2>Analogy</h2>
<ul>
<li>It takes me 3 minutes to sing Pop Goes The Weasel. I will repeat it at infinitam.</li>
<li>It takes me 20 minutes to walk around the block. I will never stop walking around the block.</li>
</ul>
<p>Therefore, when I do a full lap of the block and reach the position I started from, I will not be starting the song at the same moment. At that point in time, I will have sung it 6 times (= 18 minutes) and be 2 minutes into the 7th time singing the song (bringing us up to 20 minutes).</p>
<p>The next time I reach the initial walking position, I will have sung it 13 times in total (3x13 = 39) and be 1 minute into the 14th song.</p>
<p>The third time I reach the initial walking position, since 3 min x 20 songs = 20 min x 3 laps around the block.</p>
<p>The short answer is that the time it takes to do one thing (walk around the block / orbit the sun) is not an exact multiple of the other thing (sing a song / revolve around your own axis), therefore they don't line up neatly.</p>
<hr />
<h2>Define a &quot;year&quot;</h2>
<blockquote>
<p>if the sun appears to be in the exact point in the sky as that time the previous solar year</p>
</blockquote>
<p>If this were the definition of a year, it would inherently mean that a year is expressed as an integer amount of &quot;sun seems to revolve around the sky&quot; experiences (otherwise known as a day).</p>
<p>However, this is not how we define a year. The sun is in that same spot in the sky 365 times before we call it a year. What's wrong with the first 364 times it happens?</p>
<p>A year is not defined by the position of the sun in the sky, which is based on Earth's rotation. A year is defined by Earth's position in its orbit around the sun. Therefore the day on which the sun is in the same spot in the sky which is <em>closest</em> to that original orbital position is what we call a year, which is after 365-ish days.</p>
<hr />
<h2>Interesting tangent</h2>
<p>A human defines an Earth day as seeing the sun in the same spot in the sky.</p>
<p>If you assume the sun does not rotate around its own axis, someone living on the surface of the sun would be able to define an Earth year as seeing the Earth in the same spot in the sky.</p>
<hr />
<h2>Some simple example math, if you're keen</h2>
<p><em>I assume a year is exactly 365.25 solar days here, to make the math easy enough for an example. Finer precision is up to you to implement.</em></p>
<p>I used neat integer values in my example which means that things sometimes do line up again. In reality, however, the values are decimally complicated and it's not as easy to get things to line up.</p>
<p>We do still pick the <em>nearest</em> integer value for our calendar. The bigger loop (orbit around the sun / walk around the block) takes 365.25 times the smaller loop (Earth's rotation / singing the song), so we commonly state that a year takes 365 days.</p>
<p>Even though that's what I did in my example, in real life we don't &quot;hold out&quot; and wait for things to line up naturally. That would take 1461 years. Every year we move up .25 of a day, so we'd need 4 years to move up a single day, and it'd take 365.25 times those 4 years to catch up to an entire year.<br />
We say that a broken clock is right twice a day, but if your calendar is only accurate once every millennium and a half, that's a worthless calendar.</p>
<p>Instead we continually adjust our measurement to account for the &quot;drift&quot; (i.e. the 1 minute &quot;delay&quot; in the song every time I lap the block, or the .25 day &quot;delay&quot; in the Earth's rotation every year).</p>
<p>If we adjusted this every year, which we in theory could, then we would shift the clock (= tracks the smaller loop) by 6 hours. Our day/night cycle would shift by .25 day (6 hours) every year. Imagine having daylight savings time but the jump is now 6 hours, and during the year our clocks would drift further and further from &quot;normal&quot; until it's 6 hours out, and then it would snap back into place.</p>
<p>That would be a horrible experience.</p>
<p>Instead, we ignore the drift until we accrue a full day, which takes 4 years (4 x 0.25 = 1). When we reach that point, we jump (or should I say we leap...) a whole day.<br />
This means that we don't need to change our clocks, since we will move from midnight to midnight on a different day.</p>
<p>With this system, it only impacts our calendar, which is why we introduce this extra &quot;catch up&quot; day called a leap day.</p>
<hr />
<p>As a software developer I am perpetually astounded that we humans have an innate sense of what easiest, and it's counterintuitively one of the most horribly complex and unintuitive systems that we (developers) struggle with on a daily basis.</p>

CC BY-SA 4.0
2023-04-13T17:34:01.070Z
Surely this statement "The sun is in that same spot in the sky 365 times before we call it a year." is not true? 
Each day as we move from winter to summer, the sun rises slightly further to the east, then NE, and sets further to the west then NW. It rises higher in the middle of day, tracing an arc different to the day before. Then reverses this process as we regress back to winter. 
So surely it is only in the exact same position in the sky at a given time of day on 2 days of the year, but on those days, it is in infinitely many "places" that are the same as it was six months(ish) ago?
AdamV
CC BY-SA 4.0
2023-04-13T23:17:50.963Z
@AdamV: You're reading that statement with much greater precision than it was intended. The context is that I'm talking about the "same spot" in reference to a day cycle. The precise location is irrelevant, only that it is the same spot _in the cycle_.
Flater
CC BY-SA 4.0
2023-04-11T23:58:43.833Z
2023-04-12T00:32:07.583Z
2023-04-12T00:32:07.583Z
Flater
<blockquote>
<p>However, if the sun appears to be in the exact point in the sky as
that time the previous solar year, how does the earth reach this exact
position while only being 5hrs, 48 min, 45 secs into its final
rotation of the year?</p>
</blockquote>
<p>It isn't in the exact same point and the earth is still 5hrs, 48 min, and 45 secs behind on its track around the sun.</p>
<p>You probably feel that something is amiss, because you need a nice rigorous definition of a year. Let's give a shot at making one up.</p>
<p>Let's stand somewhere on the equator and record the sun's maximum elevation every day. If we keep doing this for a few years, we'll find that we have recorded a function with a period of 365 days, 5 hours, 48 minutes, and 45 secs.</p>
<p>Now this may feel bothersome, because from one year to another we are unable to get the sun in exactly the same position that we observed it in in the previous year. But we can cheat a bit.</p>
<p>Suppose we record the sun's position in the sky at noon (when the sun is the highest). Then we come back to the same place on the equator exactly 365 days later. We'll see, that the sun's position is ever so slightly off. But if we travel approximately 10'000km westwards (which is very close to the rotation of the earth over a period of 5hrs 48mins, etc) and wait until noon there, we'll see a much better match of the sun's elevation with what we observed a year before.</p>
<p>PS For what it's worth, the definition of a year I just made up is closest to the <a href="https://en.wikipedia.org/wiki/Year#Sidereal,_tropical,_and_anomalistic_years" rel="nofollow noreferrer">tropical year</a>. This is the one that mattered the most to our ancestors, since it lines up with the periodicity of the seasons. This link to wikipedia also discusses the sidereal and anomalistic years, which for various reasons are slightly longer.</p>

CC BY-SA 4.0
2023-04-13T14:30:03.097Z
2023-04-14T06:55:41.970Z
2023-04-14T06:55:41.970Z
LLlAMnYP
<p>Your confusion seems to stem from there being several definitions of the word &quot;year&quot;. If someone says &quot;This happened exactly one year ago&quot;, they generally mean that it happened at the exact same time of day, on the same calendar date, last year. However, the term &quot;year&quot; can also mean the time that it takes for the Earth to make one orbit around the Sun. The former is a calendar year[1]. For the latter, there are slightly different definitions that give slightly different lengths, such as sidereal, solar/tropical, and Julian, but all of them are close enough to not matter for this question.</p>
<p>The solar year is about 365 days and 6 hours long. The calendar year, however, by its very definition is a whole number of days; if it's the same time of day, then there must have been an integer number of days in between. So since the calendar year and solar year are slightly different length, we need an adjustment to keep them from getting out of sync, and that adjustment is leap days. Usually the calendar year is 365 days long, but sometimes it's 366 days long. Thus, by changing the length of the calendar year, we can have a year that is always a whole number of years have an <em>average</em> length that's not a whole number of days.</p>
<p>[1] The term &quot;calendar year&quot; is often used to distinguish between the period from Jan 1 to Dec 31 versus a fiscal year that has other endpoints, but here I'm using it to distinguish &quot;same date&quot; from &quot;one orbit&quot;.</p>

CC BY-SA 4.0
2023-04-29T16:09:44.563Z
2023-04-29T16:09:44.563Z
Acccumulation
<p>So far the solar day is still independent on the solar year. That's the reason not only for not-whole number of days in a year but in a quite complicated way of counting a calendar not to drift away.</p>
<p>By the way, there are three definitions of a &quot;day&quot;. One defines it as period of time between same positions of sun in the sky (ignoring elevation). The second defines it as a period of time between same positions of stars in the sky. And the third defines it as 86 400 seconds.</p>
<p>Quite simillar system is Earth - Moon where lunar &quot;day&quot; is exactly one lunar year by the second definition and infinite by the first definition. It's because Moon is tidally locked to the Earth.</p>
<p>Earth is not (yet) tidally locked with any other planet or star so there is no reason to have any synchronisation of the day to the year except for a sheer luck.</p>

CC BY-SA 4.0
2023-04-12T16:26:00.710Z
2023-04-12T16:26:00.710Z
Crowley
<p>@zwol made a good comment:</p>
<blockquote>
<p>I feel like all of the existing answers miss the point of the question, which might be better stated &quot;why isn't the solar year defined to be the time it takes for the sun to return to the same position in the sky, relative to the Earth's surface, as that time the previous solar year?&quot; (Which would necessarily be a whole number of solar days.) After all, the entire reason there's a difference between a sidereal day and a solar day is that we do use a &quot;return to the same position in the sky relative to the surface&quot; [sort of] definition for a solar day.</p>
</blockquote>
<p>The answer to <em>that</em> question is that defining a solar day in a way where each year is exactly 365 days would require us to do either one of the following:</p>
<ol>
<li>Change the definition of the second, so that there's still exactly 60 * 60 * 24 seconds in a single day</li>
<li>Redefine the day as 24 hours + 59.178 seconds (to account for the fact that a solar day is slightly longer than a sidereal year)</li>
</ol>
<p>Both of these changes would require a <em>massive</em> investment to change every single clock on the planet. Modern operating systems could probably be updated relatively easily, however mechanical clocks and legacy devices would all have to be replaced to account for the changes. Almost no one would agree to undertake such an investment just to get rid of February 29th, so this won't happen in the foreseeable future.</p>

CC BY-SA 4.0
2023-04-12T16:01:02.410Z
2023-04-12T16:01:02.410Z
JonathanReez
<p><strong>Perhaps an image like the following might help:</strong></p>
<p><a href="https://i.sstatic.net/COe5B.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/COe5B.png" alt="epicycloid" /></a></p>
<BR>
<BR>
<p>I am walking big circles around you.  At the same time, I am also spinning on my own axis.  The blue points in the graph above show the path that my nose would trace out as I spin around you.</p>
<p>Notice that I turn <em>approximately</em> eight circles on my own axis for every big circle that I walk around you.  But it is not exactly eight.  (Hey, no one says it has to be an exact multiple!)  Therefore, on the eighth circle, I end up facing you at a slightly different place from where I started.</p>

CC BY-SA 4.0
2023-04-20T17:11:13.723Z
2023-04-20T17:11:13.723Z
SlowMagic
<p>I've been trying to find an answer to this question for some time now, and I seem to be missing something. I understand that the rotation and revolution of the Earth are not synchronized. I understand that solar days and years are mean averages of the actual fluctuations over a year. And I know that solar days and years are different than sidereal ones. However, if the sun appears to be in the exact point in the sky as that time the previous solar year, how does the earth reach this exact position while only being 5hrs, 48 min, 45 secs into its final rotation of the year? Wouldn't that difference of almost 6hrs or 18hrs mean it would be an entirely different time of day than the previous year? Can someone please tell me what I'm missing her?</p>

CC BY-SA 4.0
2023-04-11T12:14:13.267Z
Because, in astronomy, [normal numbers are weird](https://xkcd.com/2707/).
Greg Burghardt
CC BY-SA 4.0
2023-04-11T19:22:20.077Z
I feel like _all_ of the existing answers miss the point of the question, which might be better stated "why isn't the solar year _defined_ to be the time it takes for the sun to return to the same position in the sky, relative to the Earth's surface, as that time the previous solar year?" (Which would necessarily be a whole number of solar days.)  After all, the entire reason there's a difference between a sidereal day and a solar day is that we _do_ use a "return to the same position in the sky relative to the surface" [sort of] definition for a solar day.
zwol
CC BY-SA 4.0
2023-04-11T20:42:35.593Z
I'm the OP. This comment above all the others addresses the root of my question. That's a much better way to state what I'm asking. But what is the answer to this question?
Jimmy Jam
CC BY-SA 4.0
2023-04-11T21:17:02.013Z
There is not a whole number of solar days in a solar year for the simple reason that 'days' and 'years' are purely human ideas, having nothing to do with astronomy, cosmology or anything of the kind.

When you've been trying to find an answer to this question and you seem to be missing something, how far did your quest get before what seemed to go wrong?
Robbie Goodwin
CC BY-SA 4.0
2023-04-12T16:35:28.997Z
@RobbieGoodwin Days and years represent durations of very real cyclical physical phenomena that would continue whether humans existed or not - I'm perplexed how you can argue that the motion of the planets has "nothing to do with astronomy". The notion of weeks is an arbitrary human invention, but days and years are not. It is not a "human idea" that the earth takes some fixed amount of time to complete one rotation or orbit.
Nuclear Hoagie
CC BY-SA 4.0
2023-04-12T21:58:41.563Z
@NuclearHoagie That's both obviously true and, in my view, wholly irrelevant to the Question, 'Why are there not a whole number of solar days in a solar year'?

Who doubts it's not a 'human idea' that the Earth takes a fixed time to complete an orbit?

Until you explain otherwise, that has nothing to do with any comparison of solar days to solar years?

How is that hard?
Robbie Goodwin
CC BY-SA 4.0
2023-04-13T01:08:27.470Z
@GregBurghardt it's hard to make me laugh out loud before my morning coffee kicks in, success! :-)
uhoh
CC BY-SA 4.0
2023-04-13T01:10:33.940Z
@zwol do you think there is room for that to be asked as a new question, linking back here for background and then going forward? Or perhaps "Is there a type of year defined as...?"
uhoh
CC BY-SA 4.0
2023-04-13T01:10:56.290Z
@JimmyJam see my comment to zwol above
uhoh
CC BY-SA 4.0
2023-04-13T13:01:13.293Z
@RobbieGoodwin The period of the earth's orbit is not evenly divisible by the period of the earth's rotation no matter what unit you choose to count the passage of time with. That ratio is not a whole number, whether humans exist or not. Solar days and solar years are defined by physical phenomena, not by arbitrary human choice. We didn't *choose* to have solar days and years be non-divisible, we *observed* it to be the case.
Nuclear Hoagie
CC BY-SA 4.0
2023-04-13T18:44:58.230Z
@Robbie Comments should be used to help clarify unclear questions, but your comments are confusing. There are several ways to define "day" and ["year"](https://en.wikipedia.org/wiki/Year#Astronomical_years). The OP asks about the ratio between the current mean solar day and the mean solar year. Admittedly, that's slightly ambiguous, since it's necessary to specify a reference frame.
PM 2Ring
CC BY-SA 4.0
2023-04-13T18:45:18.790Z
(cont) Traditionally, rotation and revolution are given with reference to the March equinox direction. However, that's still ambiguous because of the [precession of the equinoxes](https://en.wikipedia.org/wiki/Axial_precession), so one can use the current equinox "of date", or the equinox of a particular epoch, eg [J2000.0](https://en.wikipedia.org/wiki/Epoch_(astronomy)#Julian_years_and_J2000). The modern approach uses the [ICRF](https://en.wikipedia.org/wiki/International_Celestial_Reference_System_and_its_realizations).
PM 2Ring
CC BY-SA 4.0
2023-04-14T23:43:58.730Z
OP here. Let me rephrase this, and you all can confirm if I'm correct or not in my understanding: humans developed our measurement of 24hrs to equal a mean solar day. However, over the ensuing decades, humans realized that this 24hr measurement was not precise, which caused the calendar to slip by nearly 6hrs a year. That's why they started doing leap years. So my question is this: Could we not just update our measurement of the day to be more precise, like increasing our definition of a mean solar day to be slightly longer than 24hrs so that it lines up correctly with the mean solar year?
Jimmy Jam
CC BY-SA 4.0
2023-04-14T23:46:18.907Z
As I understand it, we use leap days to "add" an extra 24 hrs to our calendar every 4 years. Couldn't we just split this up and add it to each day, effectively making our definition of a mean solar day roughly 57ish seconds longer? I feel like doing this would be more logical than adding an entire extra day every four years. Am I wrong in coming to this conclusion?
Jimmy Jam
CC BY-SA 4.0
2023-04-15T07:33:12.127Z
@Jimmy No, a mean solar day is *very* close to 24 hours. And a mean tropical year is a little under 365.2422 mean solar days. If we used slightly longer days to avoid the need of a leap day then our clocks would be out of sync with the Sun. Eg, if the year (at some location) starts at midday 1st of Jan, then one year later the time is ~6 PM on a clock keeping mean solar time. But a clock keeping Jimmy time says its midday, even though it's near sunset.
PM 2Ring
CC BY-SA 4.0
2023-04-16T18:20:34.397Z
Jimmy Time, haha. I like that. Your description is helpful for me to visualize, thank you. Let me ask one more follow up question to make sure I understand this correctly. If our current time-keeping model did not include a leap day every four years, our clock/calendar would slip off of alignment with the sun by about six hours every year cumulatively (which is what happened with the Julian calendar). So with this hypothetical Jimmy clock/calendar that adjusts the mean solar day and removes the leap year, would it then slip in and out of sync with the sun in roughly four year cycles?
Jimmy Jam
CC BY-SA 4.0
2023-04-17T03:24:02.343Z
@Jimmy The Julian calendar uses a year length of 365.25 (solar) days, which gives 25 leap days per century. The Gregorian uses 365.2425, which gives us 97 leap days per 4 centuries. The Egyptians & Mayans used calendars of 365 days (the so-called vague year). Such calendars slip backwards through the seasons with a cycle of ~1460 years. See https://en.wikipedia.org/wiki/Sothic_cycle
PM 2Ring
CC BY-SA 4.0
2023-04-17T03:27:34.863Z
(cont) The Jimmy time clock / calendar would get back into sync with the Sun every 4 years if the tropical year length were exactly 365.25 solar days, but since it's slightly shorter, you still get a bit of slippage.
PM 2Ring
CC BY-SA 4.0
2023-04-20T17:50:45.910Z
at @zwol,  I've edited my answer.  "It's just not possible to choose an integer number of days between which the sun returns to *exactly* the same position."
James K
CC BY-SA 4.0
2023-04-29T17:05:57.187Z
I'm sorry I put that so risibly, stupidly wrongly. My only excuse is that I let myself be bemused by looking for a question left after Jimmy Jam's own exposition. What I so badly failed to say was not that 'days' and 'years' are human ideas but that here, the 'link' between them is a human idea. At least, my foolish Comment invoked Nuclear Hoagie's handy explanation which I doubt would otherwise have come in such useful detail.

I again apologise. I will happily revoke that first Comment if Members are sure that won't leave a gap in the logic of the Thread
Robbie Goodwin
<the-sun><time>
Why are there not a whole number of solar days in a solar year?
Jimmy Jam
# Why are there not a whole number of solar days in a solar year?

(Asked by: username_0 on 2023-04-10)

I've been trying to find an answer to this question for some time now, and I seem to be missing something. I understand that the rotation and revolution of the Earth are not synchronized. I understand that solar days and years are mean averages of the actual fluctuations over a year. And I know that solar days and years are different than sidereal ones. However, if the sun appears to be in the exact point in the sky as that time the previous solar year, how does the earth reach this exact position while only being 5hrs, 48 min, 45 secs into its final rotation of the year? Wouldn't that difference of almost 6hrs or 18hrs mean it would be an entirely different time of day than the previous year? Can someone please tell me what I'm missing her?

**Tags:** the-sun, time
> **Comments:**
> **username_1**: Because, in astronomy, [normal numbers are weird](https://xkcd.com/2707/).  
> **username_2**: I feel like _all_ of the existing answers miss the point of the question, which might be better stated "why isn't the solar year _defined_ to be the time it takes for the sun to return to the same position in the sky, relative to the Earth's surface, as that time the previous solar year?" (Which would necessarily be a whole number of solar days.)  After all, the entire reason there's a difference between a sidereal day and a solar day is that we _do_ use a "return to the same position in the sky relative to the surface" [sort of] definition for a solar day.  
> **username_0**: I'm the OP. This comment above all the others addresses the root of my question. That's a much better way to state what I'm asking. But what is the answer to this question?  
> **username_3**: There is not a whole number of solar days in a solar year for the simple reason that 'days' and 'years' are purely human ideas, having nothing to do with astronomy, cosmology or anything of the kind.  When you've been trying to find an answer to this question and you seem to be missing something, how far did your quest get before what seemed to go wrong?  
> **username_4**: @RobbieGoodwin Days and years represent durations of very real cyclical physical phenomena that would continue whether humans existed or not - I'm perplexed how you can argue that the motion of the planets has "nothing to do with astronomy". The notion of weeks is an arbitrary human invention, but days and years are not. It is not a "human idea" that the earth takes some fixed amount of time to complete one rotation or orbit.  
> **username_3**: @NuclearHoagie That's both obviously true and, in my view, wholly irrelevant to the Question, 'Why are there not a whole number of solar days in a solar year'?  Who doubts it's not a 'human idea' that the Earth takes a fixed time to complete an orbit?  Until you explain otherwise, that has nothing to do with any comparison of solar days to solar years?  How is that hard?  
> **username_5**: @GregBurghardt it's hard to make me laugh out loud before my morning coffee kicks in, success! :-)  
> **username_5**: @zwol do you think there is room for that to be asked as a new question, linking back here for background and then going forward? Or perhaps "Is there a type of year defined as...?"  
> **username_5**: @JimmyJam see my comment to zwol above  
> **username_4**: @RobbieGoodwin The period of the earth's orbit is not evenly divisible by the period of the earth's rotation no matter what unit you choose to count the passage of time with. That ratio is not a whole number, whether humans exist or not. Solar days and solar years are defined by physical phenomena, not by arbitrary human choice. We didn't *choose* to have solar days and years be non-divisible, we *observed* it to be the case.  
> **username_6**: @Robbie Comments should be used to help clarify unclear questions, but your comments are confusing. There are several ways to define "day" and ["year"](https://en.wikipedia.org/wiki/Year#Astronomical_years). The OP asks about the ratio between the current mean solar day and the mean solar year. Admittedly, that's slightly ambiguous, since it's necessary to specify a reference frame.  
> **username_6**: (cont) Traditionally, rotation and revolution are given with reference to the March equinox direction. However, that's still ambiguous because of the [precession of the equinoxes](https://en.wikipedia.org/wiki/Axial_precession), so one can use the current equinox "of date", or the equinox of a particular epoch, eg [J2000.0](https://en.wikipedia.org/wiki/Epoch_(astronomy)#Julian_years_and_J2000). The modern approach uses the [ICRF](https://en.wikipedia.org/wiki/International_Celestial_Reference_System_and_its_realizations).  
> **username_0**: OP here. Let me rephrase this, and you all can confirm if I'm correct or not in my understanding: humans developed our measurement of 24hrs to equal a mean solar day. However, over the ensuing decades, humans realized that this 24hr measurement was not precise, which caused the calendar to slip by nearly 6hrs a year. That's why they started doing leap years. So my question is this: Could we not just update our measurement of the day to be more precise, like increasing our definition of a mean solar day to be slightly longer than 24hrs so that it lines up correctly with the mean solar year?  
> **username_0**: As I understand it, we use leap days to "add" an extra 24 hrs to our calendar every 4 years. Couldn't we just split this up and add it to each day, effectively making our definition of a mean solar day roughly 57ish seconds longer? I feel like doing this would be more logical than adding an entire extra day every four years. Am I wrong in coming to this conclusion?  
> **username_6**: @Jimmy No, a mean solar day is *very* close to 24 hours. And a mean tropical year is a little under 365.2422 mean solar days. If we used slightly longer days to avoid the need of a leap day then our clocks would be out of sync with the Sun. Eg, if the year (at some location) starts at midday 1st of Jan, then one year later the time is ~6 PM on a clock keeping mean solar time. But a clock keeping Jimmy time says its midday, even though it's near sunset.  
> **username_0**: Jimmy Time, haha. I like that. Your description is helpful for me to visualize, thank you. Let me ask one more follow up question to make sure I understand this correctly. If our current time-keeping model did not include a leap day every four years, our clock/calendar would slip off of alignment with the sun by about six hours every year cumulatively (which is what happened with the Julian calendar). So with this hypothetical Jimmy clock/calendar that adjusts the mean solar day and removes the leap year, would it then slip in and out of sync with the sun in roughly four year cycles?  
> **username_6**: @Jimmy The Julian calendar uses a year length of 365.25 (solar) days, which gives 25 leap days per century. The Gregorian uses 365.2425, which gives us 97 leap days per 4 centuries. The Egyptians & Mayans used calendars of 365 days (the so-called vague year). Such calendars slip backwards through the seasons with a cycle of ~1460 years. See https://en.wikipedia.org/wiki/Sothic_cycle  
> **username_6**: (cont) The Jimmy time clock / calendar would get back into sync with the Sun every 4 years if the tropical year length were exactly 365.25 solar days, but since it's slightly shorter, you still get a bit of slippage.  
> **username_7**: at @zwol,  I've edited my answer.  "It's just not possible to choose an integer number of days between which the sun returns to *exactly* the same position."  
> **username_3**: I'm sorry I put that so risibly, stupidly wrongly. My only excuse is that I let myself be bemused by looking for a question left after Jimmy Jam's own exposition. What I so badly failed to say was not that 'days' and 'years' are human ideas but that here, the 'link' between them is a human idea. At least, my foolish Comment invoked Nuclear Hoagie's handy explanation which I doubt would otherwise have come in such useful detail.  I again apologise. I will happily revoke that first Comment if Members are sure that won't leave a gap in the logic of the Thread  

---

## 12 answers

### Answer by username_8 on 2023-04-10. Score: 28

> if the sun appears to be in the exact point in the sky as that time the previous solar year,

This statement means the sun is at (nearly) the same position *relative to the stars* that it was the previous year. Not that the sun is at the same point in the sky relative to the earth's surface.

So there are two motions: the sun across the field of stars, and the earth's direction to see things.

For the sidereal year (which is *really* close to the solar year), all we care about is where the sun has moved relative to the stars. The angle of earth's rotation so that your house is able to see the sun doesn't matter.


---

### Answer by username_9 on 2023-04-10. Score: 19

Let‘s turn your question around: Why should they be in sync?

The only mechanism that redistributes angular momentum between two celestial bodies are tidal forces (not considering general relativity and crashing the bodies against each other. Angular momentum of the whole system must be conserved). In the long term this leads to locked rotation as in the cases of moon, Mercury and between Pluto and Charon. As you probably know, tides created by sun and moon are creating a [drag on earth to go slower](https://en.wikipedia.org/wiki/Tidal_acceleration) and earth’s rotation was quicker in the past. Note that this force in the end is going to force earth into a locked rotation around sun (cf. spring tide). So we will end up with a day length of one year (if the force is strong enough compared to other influences on angular momentum like meteors entering earth atmosphere).

Simply put: There‘s a mechanism to create a resonance between day length and year length, but the day length is going to become 1 year. That it doesn't match an integer number of days at the current point in time is coincidence.

Note that [stronger resonances](https://en.wikipedia.org/wiki/Orbital_resonance) do exist between planet orbits (e.g. Jupiter‘s moons), which is like pushing gravitationally a child on a swing, but these forces do not couple to the angular momentum of the respective body and the tidal forces are much smaller in comparison.

> **Comments:**
> **username_10**: *There‘s no mechanism to create a resonance between day length and year length*, except that your previous paragraph explained that there is, it's just too weak between Sun and Earth to lead to resonance within the lifetime of the solar system.  
> **username_4**: Agree this answer seems to veer around in what the takeaway should be. It goes from "what mechanism could possibly be responsible" to "here is a mechanism that could do it and examples where it happens" and then concludes with "there is no mechanism, rotational and orbital period are unrelated", which rather contradicts everything else.  
> **username_9**: @gerrit thanks for catching my lapse. Hope I fixed it now.  

---

### Answer by username_7 on 2023-04-10. Score: 16

It isn't in the same place.

We can measure "one year" starting at any point in time, suppose we start at sunrise on January 1st at some convenient location (eg London). We know where the sun is; it is on the horizon. The time is about 8am.

One solar year later (there is a slight detail about the exact meaning of this which I shall gloss over) it is 1:48pm on January 1st, and the sun is not in the same place. Unless there has been a leap day, in which case it is about 1:48 on December 31st.

So you are totally right, after one solar year it is an entirely different time. We set our calendar to start each day at midnight and this means that if we didn't have leap days the calendar would quickly get out of sync with the seasons.

So why not *define* the solar year to be the time for the sun to reach the same place in the sky? Let's look at this in more detail. On Jan 1st 2022 at sunrise (in my location) it is 8:03:25 and the sun has an azimuth of 127.09 degrees (and is on the horizon). When is it in exactly the same position? Well not on Jan 1st 2023 (it has an azimuth of 127.13), nor on Jan 2nd (126.99) nor Dec 31st (127.28)

The sun doesn't come back to the same place after an integer number of days, it's pretty close *but not exact* In fact, what you would find is that after 4 years of 365 days each, the sun would be close to the 127.09 degrees on Jan 2nd. If you kept on going, after about 700 years, the sun would be rising at that azimuth in the middle of June!

It's just not possible to choose an integer number of days between which the sun returns to *exactly* the same position.

So the definition of the solar year is the time for the sun to get back to the same position relative to the equinox (the point in the sky where the path of the sun crosses the equator) This is the "tropical year" it doesn't have an integer number of days, but it does keep the seasons exactly aligned. If it is the start of spring, then 1000 "tropical years" later it will still be the start of spring (though a different time of day) And we then design a calendar to approximately align a variable number of days to the tropical year. This is convenient, it means (for example) farmers don't need to adjust for the shift in the calendar when planning the best time for planting.

This is not the only way of designing a calendar. Most notably the Islamic calendar strictly follows a cycle of twelve lunar months, and so a month such as Ramaḍān may occur in the northern hemisphere summer or winter.

> **Comments:**
> **username_11**: It seems to me that this is the only answer that a) understands the question and b) is correct.  
> **username_7**: Or possibly we have both misunderstood the question in the same way.  
> **username_12**: @MikeNakis but JamesK **does not** answer the question "Why are there not a whole number of solar days in a solar year?".  He just says "yes" to "Wouldn't that difference of almost 6hrs or 18hrs mean it would be an entirely different time of day than the previous year?"  

---

### Answer by username_13 on 2023-04-11. Score: 11

Take a bike and a race track. Mark one spot on the side of one of your tires.

Now place the bike on the starting line, making sure the mark is precisely at the lowest point of the wheel (where it touches the ground), precisely on the line.

Do one lap, then stop on the starting line.

The chances the mark on your tire will be aligned with the line at that point are quite small (it will happen only if the path you have taken around the track is an exact multiple of the circumference of your wheel).

(aside: on a race track the exact distance around the track will depend on where on the track you ride, so you can get very different results from one lap to the next)

The same happens for the Earth around the Sun. It spins, and a given meridian on the surface of the Earth will point towards the Sun every day (like the mark on the tire will be touching the ground each time the wheel makes one turn), and at the same time, it orbits around the Sun, like your bike doing a lap around the track. No reason for the same meridian to be towards the Sun exactly one year later.

You can see that either as "365 days later, when the reference meridian points towards the Sun, we haven't made exactly one lap around the Sun (not back in the same place on the orbit)" or "exactly a tropical year later, our reference meridian does not point towards the sun", which really are the same thing (the spot on your tire touches the ground somewhere other than the start line, or when you cross the start line, the spot on your tire is not on the ground).

This answers your specific question:

> However, if the sun appears to be in the exact point in the sky as that time the previous solar year, how does the earth reach this exact position while only being 5hrs, 48 min, 45 secs into its final rotation of the year? Wouldn't that difference of almost 6hrs or 18hrs mean it would be an entirely different time of day than the previous year?

At a given location, the Sun will **not** appear at the same exact point in the sky as that time the previous solar year. Either it appears at the same exact point in the sky as that time the previous **year** (i.e. 365 days later in most cases), but the Earth hasn't gone through a full orbit around the sun (we haven't crossed the start line on the racetrack yet), or if you are exactly one solar year later (we crossed the line), either it will be a the same place at a different time, or it will be in a different place at the same time.

So after a year we're indeed off by about 6 hours, the next year about 12, then third one, 18, and four years later, about 24 hours. So we are off by a day. Would seem we're back on our feet (our reference meridian again points towards the sun when we reach that famous spot on the orbit). But if we didn't do anything about that, after a (long) while, winter would happen in May.

So [some people determined a couple of millennia ago](https://en.wikipedia.org/wiki/Julian_calendar) that we would need to make a few adjustments now and then to the duration of a year, and introduced the leap years. So as we are off by about a day after 4 years, every 4 years the year is one day longer, and we're back on our feet.

Of course [some other people noticed a few centuries later](https://en.wikipedia.org/wiki/Gregorian_calendar) that this was not exact (it's not precisely 365.25 days), and more adjustments were made (so we skip leap years on years that are multiples of 100, but not multiples of 400). Still not exact either, but a lot closer.

Guess what, the minor difference between those two actually resulted in a difference of 13 days between the two calendars, and that introduced quite a bit of confusion for quite a while as not everybody switched calendars at the same time.

As a parallel, it may be interesting to note that some cultures use calendars based on the Moon rather than the Sun. Here again, no reason for the orbit of the Earth around the Sun (a year) to be a multiple of the orbit of the Moon around the Earth (about 29.5 days -- again, not a multiple of the time it takes the Earth to rotate around its axis either). This has the consequence that such calendars either need to add some extra days to "re-align" on the year, or they just drift year after year (which explains why the month of Ramadan in the Islamic Calendar occurs in March/April this year, while back in 2011 it started in August).


---

### Answer by username_14 on 2023-04-10. Score: 5

The motion of the sun in the sky doesn't exactly repeat at one-year intervals, but it's still reasonable to say that it's on a one-year cycle.

If you record the sun's position at the same time each day, the points will appear to lie on a figure-8 curve ([analemma](https://en.wikipedia.org/wiki/Analemma)). If you parametrize the continuous curve by a real number that runs from 0 to 365.25, and your first measurement is at 0, then the points you plot will be 0, 1, 2, ..., 364, 365, 0.75, 1.75, ..., 363.75, 364.75, 0.5, 1.5, .... This repeats after four years, but of course if I'd used a more accurate estimate of the year length it wouldn't. More importantly, it *doesn't* repeat after *one* year. But it is clearly cyclic with a period of one year, nevertheless.


---

### Answer by username_15 on 2023-04-10. Score: 5

We do have one planet in our solar system with a neat [orbital resonance](https://en.wikipedia.org/wiki/Orbital_resonance): [Mercury](https://en.wikipedia.org/wiki/Mercury_(planet)), with an orbital period of 87.9691 Earth days, and a sidereal rotation period of 58.646 days, which works out to almost exactly 1.5 sideral days, or 0.5 solar day, in its year. Being the closest planet to the Sun, it experiences a stronger gravitational pull than the other planets, thus the most affected by [tidal locking](https://en.wikipedia.org/wiki/Tidal_locking).

But Earth is farther away, and has its rotation affected by competing factors that Mercury doesn't have: a large Moon with its own gravitational pull, atmospheric pressure, and ocean currents. The length of Earth's day [isn't even constant](https://en.wikipedia.org/wiki/Day_length_fluctuations), and the rotation is gradually slowing down, so even if there ever were a whole number of days in a year, that wouldn't remain true over the course of millennia.


---

### Answer by username_16 on 2023-04-11. Score: 5

## Analogy

* It takes me 3 minutes to sing Pop Goes The Weasel. I will repeat it at infinitam.
* It takes me 20 minutes to walk around the block. I will never stop walking around the block.

Therefore, when I do a full lap of the block and reach the position I started from, I will not be starting the song at the same moment. At that point in time, I will have sung it 6 times (= 18 minutes) and be 2 minutes into the 7th time singing the song (bringing us up to 20 minutes).

The next time I reach the initial walking position, I will have sung it 13 times in total (3x13 = 39) and be 1 minute into the 14th song.

The third time I reach the initial walking position, since 3 min x 20 songs = 20 min x 3 laps around the block.

The short answer is that the time it takes to do one thing (walk around the block / orbit the sun) is not an exact multiple of the other thing (sing a song / revolve around your own axis), therefore they don't line up neatly.

---

## Define a "year"

> if the sun appears to be in the exact point in the sky as that time the previous solar year

If this were the definition of a year, it would inherently mean that a year is expressed as an integer amount of "sun seems to revolve around the sky" experiences (otherwise known as a day).

However, this is not how we define a year. The sun is in that same spot in the sky 365 times before we call it a year. What's wrong with the first 364 times it happens?

A year is not defined by the position of the sun in the sky, which is based on Earth's rotation. A year is defined by Earth's position in its orbit around the sun. Therefore the day on which the sun is in the same spot in the sky which is *closest* to that original orbital position is what we call a year, which is after 365-ish days.

---

## Interesting tangent

A human defines an Earth day as seeing the sun in the same spot in the sky.

If you assume the sun does not rotate around its own axis, someone living on the surface of the sun would be able to define an Earth year as seeing the Earth in the same spot in the sky.

---

## Some simple example math, if you're keen

*I assume a year is exactly 365.25 solar days here, to make the math easy enough for an example. Finer precision is up to you to implement.*

I used neat integer values in my example which means that things sometimes do line up again. In reality, however, the values are decimally complicated and it's not as easy to get things to line up.

We do still pick the *nearest* integer value for our calendar. The bigger loop (orbit around the sun / walk around the block) takes 365.25 times the smaller loop (Earth's rotation / singing the song), so we commonly state that a year takes 365 days.

Even though that's what I did in my example, in real life we don't "hold out" and wait for things to line up naturally. That would take 1461 years. Every year we move up .25 of a day, so we'd need 4 years to move up a single day, and it'd take 365.25 times those 4 years to catch up to an entire year.  
We say that a broken clock is right twice a day, but if your calendar is only accurate once every millennium and a half, that's a worthless calendar.

Instead we continually adjust our measurement to account for the "drift" (i.e. the 1 minute "delay" in the song every time I lap the block, or the .25 day "delay" in the Earth's rotation every year).

If we adjusted this every year, which we in theory could, then we would shift the clock (= tracks the smaller loop) by 6 hours. Our day/night cycle would shift by .25 day (6 hours) every year. Imagine having daylight savings time but the jump is now 6 hours, and during the year our clocks would drift further and further from "normal" until it's 6 hours out, and then it would snap back into place.

That would be a horrible experience.

Instead, we ignore the drift until we accrue a full day, which takes 4 years (4 x 0.25 = 1). When we reach that point, we jump (or should I say we leap...) a whole day.  
This means that we don't need to change our clocks, since we will move from midnight to midnight on a different day.

With this system, it only impacts our calendar, which is why we introduce this extra "catch up" day called a leap day.

---

As a software developer I am perpetually astounded that we humans have an innate sense of what easiest, and it's counterintuitively one of the most horribly complex and unintuitive systems that we (developers) struggle with on a daily basis.

> **Comments:**
> **username_17**: Surely this statement "The sun is in that same spot in the sky 365 times before we call it a year." is not true?  Each day as we move from winter to summer, the sun rises slightly further to the east, then NE, and sets further to the west then NW. It rises higher in the middle of day, tracing an arc different to the day before. Then reverses this process as we regress back to winter.  So surely it is only in the exact same position in the sky at a given time of day on 2 days of the year, but on those days, it is in infinitely many "places" that are the same as it was six months(ish) ago?  
> **username_16**: @AdamV: You're reading that statement with much greater precision than it was intended. The context is that I'm talking about the "same spot" in reference to a day cycle. The precise location is irrelevant, only that it is the same spot _in the cycle_.  

---

### Answer by username_18 on 2023-04-13. Score: 2

> However, if the sun appears to be in the exact point in the sky as
> that time the previous solar year, how does the earth reach this exact
> position while only being 5hrs, 48 min, 45 secs into its final
> rotation of the year?

It isn't in the exact same point and the earth is still 5hrs, 48 min, and 45 secs behind on its track around the sun.

You probably feel that something is amiss, because you need a nice rigorous definition of a year. Let's give a shot at making one up.

Let's stand somewhere on the equator and record the sun's maximum elevation every day. If we keep doing this for a few years, we'll find that we have recorded a function with a period of 365 days, 5 hours, 48 minutes, and 45 secs.

Now this may feel bothersome, because from one year to another we are unable to get the sun in exactly the same position that we observed it in in the previous year. But we can cheat a bit.

Suppose we record the sun's position in the sky at noon (when the sun is the highest). Then we come back to the same place on the equator exactly 365 days later. We'll see, that the sun's position is ever so slightly off. But if we travel approximately 10'000km westwards (which is very close to the rotation of the earth over a period of 5hrs 48mins, etc) and wait until noon there, we'll see a much better match of the sun's elevation with what we observed a year before.

PS For what it's worth, the definition of a year I just made up is closest to the [tropical year](https://en.wikipedia.org/wiki/Year#Sidereal,_tropical,_and_anomalistic_years). This is the one that mattered the most to our ancestors, since it lines up with the periodicity of the seasons. This link to wikipedia also discusses the sidereal and anomalistic years, which for various reasons are slightly longer.


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### Answer by username_19 on 2023-04-29. Score: 1

Your confusion seems to stem from there being several definitions of the word "year". If someone says "This happened exactly one year ago", they generally mean that it happened at the exact same time of day, on the same calendar date, last year. However, the term "year" can also mean the time that it takes for the Earth to make one orbit around the Sun. The former is a calendar year[1]. For the latter, there are slightly different definitions that give slightly different lengths, such as sidereal, solar/tropical, and Julian, but all of them are close enough to not matter for this question.

The solar year is about 365 days and 6 hours long. The calendar year, however, by its very definition is a whole number of days; if it's the same time of day, then there must have been an integer number of days in between. So since the calendar year and solar year are slightly different length, we need an adjustment to keep them from getting out of sync, and that adjustment is leap days. Usually the calendar year is 365 days long, but sometimes it's 366 days long. Thus, by changing the length of the calendar year, we can have a year that is always a whole number of years have an *average* length that's not a whole number of days.

[1] The term "calendar year" is often used to distinguish between the period from Jan 1 to Dec 31 versus a fiscal year that has other endpoints, but here I'm using it to distinguish "same date" from "one orbit".


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### Answer by username_20 on 2023-04-12. Score: 0

So far the solar day is still independent on the solar year. That's the reason not only for not-whole number of days in a year but in a quite complicated way of counting a calendar not to drift away.

By the way, there are three definitions of a "day". One defines it as period of time between same positions of sun in the sky (ignoring elevation). The second defines it as a period of time between same positions of stars in the sky. And the third defines it as 86 400 seconds.

Quite simillar system is Earth - Moon where lunar "day" is exactly one lunar year by the second definition and infinite by the first definition. It's because Moon is tidally locked to the Earth.

Earth is not (yet) tidally locked with any other planet or star so there is no reason to have any synchronisation of the day to the year except for a sheer luck.


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### Answer by username_22 on 2023-04-20. Score: 0

**Perhaps an image like the following might help:**

[![epicycloid](https://i.sstatic.net/COe5B.png)](https://i.sstatic.net/COe5B.png)

  
  

I am walking big circles around you. At the same time, I am also spinning on my own axis. The blue points in the graph above show the path that my nose would trace out as I spin around you.

Notice that I turn *approximately* eight circles on my own axis for every big circle that I walk around you. But it is not exactly eight. (Hey, no one says it has to be an exact multiple!) Therefore, on the eighth circle, I end up facing you at a slightly different place from where I started.


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### Answer by username_21 on 2023-04-12. Score: -1

@zwol made a good comment:

> I feel like all of the existing answers miss the point of the question, which might be better stated "why isn't the solar year defined to be the time it takes for the sun to return to the same position in the sky, relative to the Earth's surface, as that time the previous solar year?" (Which would necessarily be a whole number of solar days.) After all, the entire reason there's a difference between a sidereal day and a solar day is that we do use a "return to the same position in the sky relative to the surface" [sort of] definition for a solar day.

The answer to *that* question is that defining a solar day in a way where each year is exactly 365 days would require us to do either one of the following:

1. Change the definition of the second, so that there's still exactly 60 \* 60 \* 24 seconds in a single day
2. Redefine the day as 24 hours + 59.178 seconds (to account for the fact that a solar day is slightly longer than a sidereal year)

Both of these changes would require a *massive* investment to change every single clock on the planet. Modern operating systems could probably be updated relatively easily, however mechanical clocks and legacy devices would all have to be replaced to account for the changes. Almost no one would agree to undertake such an investment just to get rid of February 29th, so this won't happen in the foreseeable future.


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