Re: Sci.astro.research experiment for astro.amateur

From: Oriel36 (geraldkelleher_at_hotmail.com)
Date: 06/17/04 

Date: 17 Jun 2004 11:35:40 -0700

No new postings have come up on the google newsreader so this will
have to do.By right it should be easy stuff to understand the error
Flamsteed created by linking the Earth's axial rotation directly to
stellar circumpolar motion.

The sidereal value of 23 hours 56 min 04 sec is based on the constant
axial rotation of the Earth through 360 degrees in 24 hours
exactly.That 24 hour/360 degree value is derived from the Sun based
reference and the Equation of Time adjustment for constant axial
rotation.

Chris L Peterson <clp@alumni.caltech.edu> wrote in message news:<9gcuc05iklblqccmle0u7dmqlokk9a47bo@4ax.com>...
> On 15 Jun 2004 09:42:15 -0700, geraldkelleher@hotmail.com (Oriel36) wrote:
>
> >An amateur astronomer using only a stopwatch can affirm that after two
> >revolutions of the Earth through 360 degrees,a star will NOT align
> >back to the same position after 47 hours 52 min 08 sec ( 23 hours 56
> >min 04 sec X 2 )...
>
> Of course it won't. Because of the tilt of the axis, it's altitude will change.
> But if you do the experiment correctly- look at the time a star crosses the
> meridian one night and on subsequent nights, you will see the length of the
> sidereal day quite clearly.
>
> For example, at my location Regulus will cross the meridian tonight at 17:32:27.
> In two nights, it will cross the meridian at 17:24:35. The difference is
> 00:07:52. Half of that is 00:03:56. Subtract that from 24 and you get 23:56:04,
> the length of the sidereal day.
>

Flamsteed calculated the time for the transit of a star back to his
meridian as 23 hours 56 min 04 sec using the 24 hour/360 degree
equivalency.He then started from scratch as the star crossed the
meridian and using the same 24 hour/360 degree equivalency for the
axial rotation of the Earth he calculated the next 23 hours 56 min 04
sec value.The reason that you are getting the 3 min 56 sec difference
is because the sidereal value is based on the constant axial rotation
of the Earth through 360 degrees in 24 hours exactly.

Unless you wish to stop the Earth from rotating 3 min 56 sec each 24
hour day,I assure you that you cannot prove constant axial rotation
via Flamsteed's sidereal method insofar as that concept is based on
the already existing assumption of constant axial rotation through the
24 hour/360 degree longitude equivalency.

 
> Okay, I didn't get those times with a stopwatch. I calculated them using the
> standard (simple) formulas, which of course take into consideration the length
> of the sidereal day. Are you saying those formulas are not accurate?

Flamsteed would have needed to check his pendulum clock at his
location each noon to determine if the clock lost or gained
seconds,this is in the absence of the accurate clocks developed later
by John Harrison for solving the longitude problem.There were no
timezones in Flamsteed's era nor did the astronomical day begin at
midnight,in this respect the 24 hour/360 degree longitudinal
equivalency is based entirely on the axial rotation of the Earth back
to the noon determination.As there is a natural inequality for each
axial rotation,the Equation of Time adjustment was applied to equalise
the natural variation in the noon determination which facilitates the
seamless transition from one equable 24 hour day to the next.

This being so,axial rotation was already known to be constant but
Flamsteed transfered it to a geocentric 'mean sun' term.It created a
'mean sun' fiction which displaced the actual assumption of constant
axial rotation via the 24 hour/360 degree equivalency,he imposed a
constant orbital motion for the Earth which is in direct conflict with
Kepler's second law,he has constant axial rotation to the Sun every 24
hours which is in direct conflict with the neccesity of the Equation
of Time,he transfered an axial coordinate to a .986 degree/3 min 56
sec orbital coordinate so how much damage can one man do.

http://astrosun2.astro.cornell.edu/academics/courses//astro201/sidereal.htm

> Interesting, because night after night I use them to aim my telescope at imaging
> targets, and night after night I find those targets dead center in my field of
> view. That's about as empirical as you can get.
>

Look,I am still searching for a proper term to describe Flamsteed's
maneuvering,normally I would just let men figure it out themselves and
use their best judgement to correct Flamsteed's astronomical vandalism
but perhaps the damage has gone on for so long (1677),it may now be
beyond recovery.Flamsteed's maneuvering can be condensed into a simple
format and if men remained convinced that the Earth rotates through
360 degrees in 23 hours 56 min 04 sec they can enjoy what that looks
like from a heliocentric view,where the Earth can swop positions with
the Sun,where there is no Kepler's second law,where the Earth not only
has a constant axial rotation wrt to the other stars but our local
star as well.

360 degrees = 24 hours

1 degree = 4 min

.986 degree = 3 min 56 sec

24 hours minus 3 min 56 sec = 23 hours 56 min 04 sec

http://astrosun2.astro.cornell.edu/academics/courses//astro201/sidereal.htm

 
> The concept of the sidereal day is not "theory" or something to be proved or
> disproved. It is simply geometry. Your continued failure to grasp this most
> trivial of concepts is rather astonishing. Of course, the value is only a mean.

I recognise that you are not insulting people's intelligence and that
you genuinely like your celestial sphere, favorite constellations and
what-you-see -is-what-you-get type sidereal reasoning.The sidereal
value is great for cataloguing and the civil calendar but it is
ultimately based on the equable 24 hour day and the constant axial
rotation of the Earth through 360 degrees and the Equation of Time
correction.

> In reality the length of the day is not constant, a fact that is recognized,
> measured, and modeled. This inconstancy is the result of gravitational effects
> and the fact that the Earth is not a perfectly rigid, uniform body. But those
> effects on timing are exceedingly tiny compared to the very obvious sidereal day
> which, contrary to your assertion, anyone with a stopwatch (and, I should add, a
> modicum of intelligence and common sense) can easily measure over the course of
> one or two days.
>

http://go.owu.edu/~jbkrygie/krygier_html/geog_222/geog_222_lo/geog_222_lo12_gr/longitude.jpg

I have explained as far and as best I can why it is a mathematical and
observational certainty that it will not,if you choose to believe it
does,I leave you with the heliocentric justification for your choice,a
choice where Kepler's planetary laws do not exist and the exquisite
Equation of Time adjustment which facilitates the equable 24 hour day
is destroyed.

http://astrosun2.astro.cornell.edu/academics/courses//astro201/sidereal.htm

> _________________________________________________
>
> Chris L Peterson
> Cloudbait Observatory
> http://www.cloudbait.com

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