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The earth completes approximately 366 rotations per year, it completes
one rotation in 23 hours-56 minutes-3.9 seconds.

gl

On Jul 18, 12:18 am, "Greg R."  wrote:
> --- "Gary J. LaPook"  wrote:
>
> > It's actually 15.041� per hour (15� 2.5') approximately 361� per
> > solar day.
>
> Hmmm... not questioning your math, but if the Earth rotates 1� beyond a
> complete rotation every day, wouldn't we need to add leap days every
> year (i.e. 365� "extra" rotation in a year = 1 extra day + 5� "left
> over"), instead of approx. every 4? Seems like it should be closer to
> something like 360.25�/day (?).
>
> Then again, it's late and I'm not thinking clearly on this one....
>
> --
> GregR
>
> --- "Gary J. LaPook"  wrote:
>
> > Gary writes:
>
> > It's actually 15.041� per hour (15� 2.5') approximately 361� per
> > solar day.
>
> > gl
>
> > Bill wrote:
>
> > As understand it, with an earth rotation of 15d per hour, 1 second
> > time
> > equals 0.25 arc minute.  It follows that 4 seconds time would equate
> > to 1
> > arc minute.
>
> > >Bill asked
>
> > >>>What time is it, really?
>
> > >>I believe the musical group Chicago answered that question back in
> > the
> > >>late '60s... ;-)
>
> > >And does anyone really care?  I do.
>
> > >>>A while ago there was a thread on time and the affect of dropping
> > >>>leap seconds on cel nav.
>
> > >>Don't think I was on the list for that thread, but as I understand
> > it
> > >>leap seconds are added to UTC as needed to keep it within 0.9
> > seconds
> > >>of astronomical time.
>
> > >>The rule that I remember from back when I was first learning celnav
> > was
> > >>that your observation time had to be accurate within 4 seconds,
> > >>otherwise your LOP could be off by up to 1 NM just from that error
> > >>alone (I interpret that to mean +/- 2 seconds). So I would say that
> > >>unless you need exceptional accuracy with your celnav sights you're
> > >>probably OK just ignoring the leap seconds.
>
> > >As understand it, with an earth rotation of 15d per hour, 1 second
> > time
> > >equals 0.25 arc minute.  It follows that 4 seconds time would equate
> > to 1
> > >arc minute.  An arc minute of longitude would be nominally 1 nm at
> > the
> > >equator, but less if the vessel's AP is north or south of the
> > equator.
> > >Roughly 1' longitude * cos latitude = fraction of a nautical mile
> > (ignoring
> > >oblateness).  For example, near an elevated pole 360d longitude
> > could be
> > >under 1 nautical mile.
>
> > >And why--despite the "former" CTA's cavalier attitude towards
> > >chronometers--would I care?  With an artificial horizon, my Astra,
> > and a 3.5
> > >scope, I consider an intercept of 0!0 from an average of 5 or more
> > >observations from a known GPS position lucky. 0!1 very good.  0!2
> > average.
> > >0!3 fair, and > 0!3 has me checking IC and sextant calibration.
>
> > >I figure an artificial horizon cuts IE and observation errors in
> > half, so it
> > >gives me 0!0 to 0!6 (averaged-observations intercept) as goal to
> > shoot for
> > >under ideal conditions.
>
> > >I have never experienced my ideal conditions.  They would include a
> > crisp
> > >horizon, clear sky, and a relatively stable (or predictable)
> > platform. And
> > >of course accurate UT1 time.  But if I ever do...
>
> > >Bill B
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