Measuring How Many Days Are in a Year - Comments

[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

 

[Jim60]

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.

 

[Jim60]

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.

 

[Jim60]

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

 

[mfb]

Where is the point in collecting all those numbers?

 

[Jim60]

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?

 

[mfb]

Maybe I need more than 25 years?

Depends purely on the precision you want to achieve.

 

[Jim60]

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.

 

[mfb]

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.

 

[Jim60]

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?

 

[Jim60]

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?