Insights Measuring How Many Days Are in a Year - Comments

[SteamKing]

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:

https://en.wikipedia.org/wiki/Equatorial_ring

 

[mfb]

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.

 

[Jim60]

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.

 

[Jim60]

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?

 

[Drakkith]

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.

 

[Jim60]

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.

 

[Drakkith]

Orienting it would be a delicate process, but offhand I would think you could set one up just about anywhere.

 

[Jim60]

Checking how many days from equinox to equinox using this site, I’m puzzled that the time varies from year to year.

http://aa.usno.navy.mil/data/docs/EarthSeasons.php

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.

 

[mfb]

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

 

[Jim60]

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?

 

[Jim60]

Did you notice on that site that the Solstices, Equinoxes and Perihelion to Perihelion varied quite a lot from the norm?

 

[anorlunda]

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.

 

[Jim60]

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.

 

[Jim60]

I should have put UTC after minutes, silly me.

 

[mfb]

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.

 

[Jim60]

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?

 

[Vanadium 50]

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.

 

[jim mcnamara]

Time lapse exposures of the analemma:

https://www.google.com/imgres?imgur...ved=0ahUKEwi62ISM56fMAhUL9WMKHeG4CGIQ9QEIIzAA

 

[mfb]

You make sound easy?

It is easy.
Manhattan has great reference points and good horizon sight with the street layout, the effect is quite famous there. 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.

 

[Jim60]

I’m getting confused here, what year are we actually measuring? Is it the sidereal year, or the Tropical year?

 

[mfb]

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

 

[SteamKing]

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.

https://en.wikipedia.org/wiki/Tropical_year
https://en.wikipedia.org/wiki/Sidereal_year

 

[mfb]

Thanks. Wanted to write days of course, but somehow messed it up.

 

[Jim60]

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.

https://en.wikipedia.org/wiki/Tropical_year
https://en.wikipedia.org/wiki/Sidereal_year

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

 

[D H]

Checking how many days from equinox to equinox using this site, I’m puzzled that the time varies from year to year.

http://aa.usno.navy.mil/data/docs/EarthSeasons.php

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.

 

[D H]

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.

 

[Old Tele man]

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.

 

[Jim60]

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?

 

[Jim60]

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

 

[PAllen]

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