[time-nuts] How many seconds in a year?
Brooke Clarke
brooke at pacific.net
Wed Oct 15 15:19:42 UTC 2008
Hi Neville:
Leap Years do not occur once every 4 years but are on a more complex schedule.
There's an exception for years that are exact multiples of 100 years.
There's an exception to the exception for years that are exact multiples of 400
years. So on a very long term average there are 365.2425 days per year. See:
http://en.wikipedia.org/wiki/Leapyear
There have been proposals to add a 4000 or 8000 year exception, but that's so
far in the future and the Earth's rotation has enough variation that these have
not been made part of the leap year calculations.
The Earth used to be the definition of time but once it was demonstrated that
Cesium clocks kept much better time they became the definition of the time. At
first the length of the second was going to be changed each year "rubber
seconds", but that was soon discarded and replaced with the leap second.
There are a couple of related areas. If you have read about or seen the movie
about Harrison, his clocks and the longitude you know part of the story.
The other is the search for the latitude. This turns out to be a much more
complex problem and took a lot longer before there was an understanding of it.
After I moved to Ukiah, CA I discovered there was a street called Observatory
Ave. Doing some research at the local Historical house determined it was one
of six of the Latitude Observatories all located very close to 39:08 deg N.
http://www.prc68.com/I/UkiahObs.shtml
Have Fun,
Brooke Clarke
http://www.prc68.com/P/Prod.html Products I make and sell
http://www.prc68.com/Alpha.shtml All my web pages listed based on html name
http://www.PRC68.com
http://www.precisionclock.com
http://www.prc68.com/I/WebCam2.shtml 24/7 Sky-Weather-Astronomy Web Cam
Neville Michie wrote:
> Thanks Jim,
> you told me what I wanted.
> We can listen to WWV to hear the stroke of midnight each day, and we
> know it is UTC and
> subject to the equation of time but otherwise it represents an exact
> rotation of the planet. We know a day is about
> 86400.00X seconds long, but a year is rounded to the nearest day,
> what is its real length, as
> it is not an integral multiple of days.
> Now 12:00:00 is only the end of the civil and legal year and there
> are errors of around 6 hours
> each year in rounding it off to the nearest day.
> Jim has given the length of a real year (the inverse of the frequency
> of the planet orbit) and
> I would be pleased to ask him another question as to the phase of
> the annual turnover,
> what time of day the next annual solstice? occurs.
> That is the time for my new year chime to sound,
> and Jim has already given the number of seconds to the next chime.
> I was not really thinking of leaving an unattended clock running for
> 200 years, but it would be nice
> to think that I had used the precision available to observe the real
> milestones in time.
> I am sorry that this question has been so difficult to ask, but there
> are so many arbitrary distortions
> of the observation of time that it has become a very fuzzy concept.
> cheers, Neville Michie
>
>
>
> On 15/10/2008, at 4:19 PM, Jim Palfreyman wrote:
>
>> Hi Neville,
>>
>> There are a few inaccuracies in your summary but the figure you
>> want for
>> J2000.0 is 365.242 190 419 SI days=31556925.252202 SI seconds.
>> However we
>> cannot give you an average over the next 200 years because it is
>> steadily
>> dropping by 5.324 ms per year. So in 200 years time it will be a
>> second out.
>>
>> However, for your purposes if you use the number above you will get
>> very
>> close to what you want.
>>
>> Regards,
>>
>> Jim
>>
>>
>> 2008/10/15 Neville Michie <namichie at gmail.com>
>>
>>> I have been thinking about my problem, the major part of the problem
>>> is finding the right question.
>>> If leap seconds are applied to keep the meanderings of the planet in
>>> phase with our mean time
>>> clocks, then what about our leap year days which are applied to keep
>>> the seasons in phase with our
>>> calendars? Applying a whole day every so many years may keep the
>>> civil authorities happy,
>>> but when we all celebrate a new year we are not counting down the
>>> completion of a planetary cycle
>>> but the civil approximation to a new year.
>>> Now, having thought about it a bit more, the year is the completing
>>> of an orbit of our elliptical orbit around the sun.
>>> The start and finish of an orbit is defined by the inclined axis of
>>> rotation of the Earth crossing the plane of its orbit which is what
>>> causes the seasons.
>>> The elliptical orbit is moving, hence the precession of the
>>> equinoxes. As we can not measure a year
>>> by looking at a particular star crossing our meridian,
>>> we measure the year by watching a fictitious star crossing our
>>> meridian. This star (which used to be at the first point of Aries
>>> on the vernal equinox) is slowly orbiting around our solar system as
>>> our tilted ecliptic precesses.
>>>
>>> Which now gets me back to my question. If I want to make a clock that
>>> chimes once per solar year when we have completed a cycle it will
>>> count down a
>>> number of seconds then chime. So I want to know the number of seconds
>>> in a Mean Year, accurate enough so that in 200 years time
>>> it will still be right.
>>> It should be fun to see the time the new year really starts as we
>>> head around the orbit again. The time will not be midnight
>>> but will jump around in the day with different years.
>>> There are so many interference's in time keeping by accountants and
>>> politicians I was thinking of making a clock that
>>> showed real time as determined by sun and stars.
>>> Although that brings up the equation of time, and mean time is so
>>> convenient if you have an atomic clock.
>>> I might just try to make a clock that shows the mediaeval "Italian
>>> Hours", where every day has 24 hours with sunrise at 6 AM with sunset
>>> at 6PM.
>>> The old clocks had dials marked so that you read off the elastic
>>> Italian hours.
>>> cheers, Neville Michie
>>>
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