# NavList:

## A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding

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Re: Lunars: Jupiter's BIG.
From: Trevor Kenchington
Date: 2003 Dec 23, 18:18 +0000

```Fred Hebard wrote:

>   One might not be able to rate a chronometer to the second using
> lunars, but it could be rated to the minute, or about 30 seconds in the
> hands of a good practitioner.

I can see that knowing the difference between GMT and chronometer time
to the nearest 30 seconds would be useful -- less than ideal, of course,
but enough to get your longitude to within ten miles.

But what benefit would there be in knowing the chronometer's rate to the
nearest 30 seconds? (And what does that mean anyway: 30 seconds per day?
Per week?) If a navigator was to take two lunar-distance observations,
two days apart (with a third on the day between as a check, if desired)
and if each observation had 95% confidence limits of 30 seconds either
side of the best estimate, then the estimate of the time elapsed between
those two observations would have 95% confidence limits of about 14
seconds (I think), which would mean 7 seconds per day when compared to
the chronometer's going. After a week without another lunar, the
navigator would not know GMT to better than somewhere in a 100-second
band, which translates into up to a 25-mile long area of possible
position. After a month, that would swell to 100 miles, in
near-equatorial latitudes. That is hardly a useful level of precision.

A navigator who became uncertain of his chronometer's rate could resort
to frequent lunars to check watch error but it would need a very long
series of them before he would know the chronometer's rate well enough
to quit doing the lunars and instead trust to the machinery for more
than a few days at a time.

Fred also wrote:

> With a regression line, it is
> important that a point near the middle of the line be chosen, and, in
> general, averaging is much better than these graphical methods for
> determining the time.  I believe George is making the same point.

I think it is true that linear regression will always plot a line
through the point defined by the average of the x values and the average
of the y. If so, picking a point on the fitted line somewhere near the
centre of the data will give a result essentially identical to averaging.

Trevor Kenchington

--
R.R.#1, Musquodoboit Harbour,                     Fax   (902) 889-9251
Nova Scotia  B0J 2L0, CANADA                      Home  (902) 889-3555

Science Serving the Fisheries

```
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