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Why is it unlikely that any bright star will be within an arc minute
of zenith?


Quoting Peter Monta :

> Hi,
>
>
> ... an easier method to locate longitude and latitude by having our
>> sightings match this knowledge via time and sighting with any of the
>> luminaries when at Zenith let's say
>>
>
> If you know the time and can see a star at the zenith, then you can indeed
> determine both your latitude and your longitude from this single event.
>
> But there are practical difficulties.  The zenith is not marked on the sky;
> you really only have the sea horizon as a reference (or spirit leveling if
> on land).  If the star is the same angular distance to each part of the
> horizon, then it is indeed at the zenith, but this takes time to verify,
> and meanwhile the star is moving.  Also, it is unlikely that any bright
> star will be within an arcminute or so of the zenith, so the available
> navigational events are severely restricted unless you can deal with very
> dim stars.
>
> So it's better to work with several of the things you can observe easily,
> namely, angular distances of various bright stars to the horizon, observed
> separately over the course of a few minutes.  A star not at the zenith
> gives only a "circle of position", not a position directly (as you don't
> know the star's azimuth to any accuracy), so you need several of these
> observations.
>
> But, as you say, computation and instrumentation are moving targets, and
> it's always interesting to see if they can map onto the classical
> navigation problems in new ways.  Sketch out a complete version of your
> zenith scheme and try it under the real sky.  (Using a camera, you can see
> dim stars very near the zenith.)  This kind of playing around can provide,
> I think, some good intuition about how the classical techniques have been
> shaped---maybe better than reading a textbook treatment of the techniques
> already fully formed.
>
> Cheers,
> Peter
>

   

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