Re: Earth Deceleration theory likely a farce

Moshiachyozif wrote:
"Greg Neill" <gneillRE@xxxxxxxxxxxxxxxx> wrote in message
news:4970d52d$0$22992$9a6e19ea@xxxxxxxxxxxxxxxxxxxxxxx
Just for fun I took the Delta-T values listed in the first column
of the second table on the web page:

http://user.online.be/felixverbelen/dt.htm

and fitted a second order polynomial to the values, first converting
the date values to "years ago" starting in year 2000. That is, year
2000 becomes t = 0, and 2000 BC becomes t = 4000.

The polynomial is of the form:

Dt(t) = a0 + a1*t + (1/2)a2*t^2 where t is "years ago"

so that the coefficients are, and can be interpreted as:

a0 = 125.98 sec : initial offset at t = 0 (year 2000)
a1 = -1.322 sec/yr : initial rate of change of offset
a2 = 5.99 msec/yr^2 : acceleration in rate of change of offset

Note that this is a very crude model that doesn't take into account
the fact that the Earth rotation rate, and the change in rotation
rate, and the change in the change of rotation rate,..., actually
fluctuate nonuniformly due to myriad physical processes. However,
that being said, it does actually fit the data values pretty well.

Now, we can differentiate the polynomial to determine the rate of
change of the Delta-T value for given years:

dD(t) = a1 + a2*t

and for 500 BC (or t = 2500 years ago) the expression gives 13.6
sec/yr change in Delta-T, corresponding roughly to a mean day length
about 37.2 msec shorter than 24 hours for that year.

Hi Greg,

Could you simply tell us what the length of the year would be in 1250 BC
if
the delta-T is 8 hours 20 minutes? This length of time is decreated to
zero by 1600 AD.

The length of the year is fixed by the Earth's orbit, and is about
365.25 days (of 24 hour days) long. Or about 31557600 seconds, if you
prefer.

Using my rough model, the value of Delta-T for 1250 BC is about 7.63 hours,
and the rate of change per day about 49.68 msec. That is, the day was about
50 milliseconds shorter back then.

So one orbit of the Earth is 365.25 x 24 hrs, and the day length in 1250 BC
about 24hr - 49.68msec. That yields the number of observed days in that
year to be:

(365.25 x 24 hrs)/(24 hrs - 49.68 msec) = 365.25021 "days"


If the earth's rotational speed was 8 hours 20 minutes faster per year,
then
could you tell us how may days in the year that would affect.

I don't know how you arrive at a rotation speed changing by that amount
per year. That would correspond to a change of over 82 seconds per day,
each day, in the value of Delta-T.


Now I'm going on the very general presumption that if the rotational speed
is increased then the length of the day is shorter. The slower the earth
rotates, the length of the day is longer. If the days are shorter then
you
can squeeze more day into the year. At the very least, the length of the
year would not be exactly the same as it is now. So what gives?

Am I right in assuming that if the earth was rotating 8 hours 20 minutes
faster in 1250 BCE than it is now that the length of the year in terms of
the number of days of the year would be more than 365.25 as it is now?

It looks to me like the Earth is rotating only about 50 milliseconds
faster in 1250 BC. Your 8 hours and 20 miniutes looks like a total
offset (Delta-T) to me.


Just tell a total lay person what difference an 8 hr 20 minute delta-T
affects the length of the year in 1250 BCE compared to the number of days
in
the year now?

It doesn't have to affect it at all if, due to the various influences,
the day happened to be 24 hours long at that time. We know it wasn't,
but the point is that Delta-T alone does not specify the day length,
it only specifies the time *offset* from a universal mean time. To arrive
at the day length you need to fit a model to the points as I did and
take the derivative of the function, as I did.

Note that my model is a crude one, as I already stated. I also left
out stating the assumption that the year 2000 day length is taken to
be a "true" 24 hour day.



.

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