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--- "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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