# NavList:

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

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Re: Calculated Altitudes for Lunars
From: Bill Noyce
Date: 2002 Oct 23, 09:19 -0400

```Herbert Prinz asks Bruce Stark:

> The one thing that leaves me dumb-founded is your remark that the
> described method would offer a 30 times improvement over "standard
> procedures" in accuracy (obtained in the first pass of iteration,
> I presume). What standard are you referring to? I thought what you
> proposed WAS more or less the standard procedure. Can you give us
> a reference for the flawed procedure you have in mind and elaborate
> on how it would give an error in GMT as big as you indicate?

If you knew your longitude but only guessed GMT, then the error in
computed LHA of any body would be about equal to the error in GMT
(converting time to arc).  If your longitude were also unknown, but
the errors in longitude and GMT were unrelated, the computed LHA
would be even larger.  And for most sights, most of the error in LHA
shows up as error in altitude.

But if you use the method Bruce describes, then even though you don't
know longitude or GMT, you do know local time (measured LAT, computed
LMT).  As a result, if your DR longitude is east of your actual
position, you will think GMT is earlier by an equal amount; if your
DR longitude is too far west, you'll think GMT is later.  As a result,
your calculated LHA for most bodies will be very good -- the two
related errors simply cancel out.  (This shouldn't be surprising for
the sun, since your LAT sight essentially established the sun's LHA.)
Stars and planets are almost as good as the sun, since their motions
are very close to 15 degrees per hour.  The moon's motion is the
one that differs most from 15 degrees/hour (which is why it's useful
for establishing time and longitude!).  Consider the moon's motion
to be the sum of two terms:  15 degrees per hour caused by the earth's
rotation, minus about 1/2 degree per hour caused by the moon's orbit,
which will add up to 360 degrees over about 28*24 hours.  The errors
in the first term cancel out just like errors in the sun, stars, and
planets, as long as our estimated GMT and estimated longitude have
equal and opposite errors (which happens if we know LAT or LMT).
So the remaining error in calculated LHA is only 1/30 as large as if
we used GMT and longitude with unrelated errors (even if one of them
is known precisely).

From working through a few examples, it looks to me as if this reduces
the error so much that iteration is not required in practical
is valid.

```
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