Re: Before the nonsense breaks out

oriel36 wrote:
.For a star to return to a meridian 3 minutes 56 seconds earlier each
24 hours and without exception requires the calendar system of 3 years/
365 days and 1 year/366 days,For any intelligent person that should be
enough ,despite the belief that the 23 hour 56 minute 04 second value
is a product of 365.24 days,the observations show it to be a
derivative of the 365/366 day calendar system.It is supposed to be a
shocking oversight,error or whatever other way you put it.

Well, let's see - the "365/366 day" calendar system in present use has 97
years of 366 days (since years divisible by 100 but not 400 are not counted
as leap years) which means that the year is considered to have an actual
length of 365 + 97/400 or 365.2425 mean solar days. Since the orbit of the
earth around the sun adds one sidereal day for each year, that means that
the length of the sidereal day is (365.2425 / 366.2425) = 0.99726956866
times the length of the mean solar day.

Since the hour is defined as 1/24th of the mean solar day, that means
that the difference is .00273043134 mean solar days, or 235.909 seconds
(3 min. 55.909 seconds).

If I perform the same calculations assuming a year length of 365.24 or 365.25
mean solar days, I then get figures of 235.911 or 235.904 seconds, respectively,
for the difference between the lengths of the solar day and the sidereal day.

All of these figures are within 0.1 second from the figure of 236 seconds
(3 min. 56 seconds) given as Flamsteed's observed difference between the
sidereal and solar day periods.

I would call this an "exact match within the resolution of available data",
since I doubt that the clocks available to Flamsteed were capable of reading
to sub-second resolution - and stating a figure down to the second probably
required the averaging of multiple observations - and careful determination
of, and compensation for, measurable variations in the clocks themselves.

Where is the evidence of "error" or "oversight" here? An observation of
the length of the sidereal day to within 1 second does not have sufficient
resolution to support a distinction between a calculated year length of
"365.24" or "365.25" days (i.e., the average year length implied by a
calendar of "3 years x 365 days + 1 year x 366 days").

-dave w
.

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