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Thank you, Tony. In other words, I could construct a solution on the univ.
plotting sheet, as I mentioned, but use the mean of departure, and
destination latitudes, and that would work? Thank you.

Ed
----- Original Message -----
From  "Tony" 
To: 
Sent: Sunday, February 27, 2000 9:00 PM
Subject: Re: Figuring Course given Lat/Long of destination


> Ed:
>
> When you say that "there is the error of the Macerator thing", can you be
> more specific?  Did you use Bowditch Mercator sailing by tables?  This
> should work out OK.
>
> Actually, just using Plane sailing with mid-latitude should be quite close
> because the distance is relatively short; only earth eccentricity is
ignored.
>
> Why the problem suggests also GC (great circle) does not make much sense.
> There would be less than a mile difference.  I did check the results by
> computer and they are OK.  [ Sometimes they are not. ;) ]
>
> Tony    in San Francisco
>
>
> > Ed Kitchin wrote:
> >
> > An interesting problem appears in the latest issue of "Ocean Navigator"
Which asks that you figure
> > the course to a destination given origination and destination. It would
seem easy to determine the
> > difference in lat. (The destination was over several degrees of lat.),
but deg. of long. differ in
> > length as you change lat. One could simply take the mean of the two
given long. and use that, but
> > that bothers me as not being all that accurate. There is the error of
the Macerator thing. You
> > could use universal plotting sheets and construct using a vertical
representing diff./lat., then
> > draw a horizontal from the top of the lat. fig., representing the long.
at the destination, and
> > draw a hypotenuse as the course line. (???) Are there any mathematicians
out there to
> >  give me a good formula to learn for this task? Thank you.
> >
> > Ed Kitchin
>

   

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