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    Re: Figuring Course given Lat/Long of destination
    From: Tony S
    Date: 2000 Feb 28, 5:02 PM

    Ed:
    
    OK on your first part. Let us know how things compare.
    
    As for what Dr. Kolbe offered I'll bow to his follow-up explanation.
    (hopefully he is watching). He is correct on what can be done on GC
    using SR tables.
    
    Tony
    
    Ed Kitchin wrote:
    >
    > Thank you, Tony. I'll check my old Bowditch and look for the tables, and
    > compare to the construction method to compare results. Meanwhile another
    > writer stated that the great circle course could be found by " using a
    > regular sight reduction table, substituting the lat./long of destination as
    > the GP of a heavenly body." He then said to "crank the handle" and get the
    > Zn as your great circle course. Now...I can do celestial nav. thanks to
    > recently taken courses using HO 249, or the electronic calculator. I am
    > trying to grasp this other guy's concept here. Seems though he is asking me
    > to work backward through the process, given that sight reduction is to
    > OBTAIN the GP, your distance off, and the Zn. Excuse my ignorance, but I
    > can't grasp how to do that. What would you use then for the Hs, and what
    > corrections would you apply? OR!!! (I just had this idea) You could enter HO
    > 249 with the arguments: lat. of destination, and long. of dest. as
    > declination, to obtain Zn - - but you would STILL need a corrected altitude
    > (Hc). I have no idea. Would you help a rank beginner out with this one?
    > Thank you.
    >
    > Ed Kitchin
    > ----- Original Message -----
    > From- "Tony" 
    > To: 
    > Sent: Monday, February 28, 2000 6:14 PM
    > Subject: Re: Figuring Course given Lat/Long of destination
    >
    > > Ed:
    > >
    > > Well, not quite. I was really encouraging you to use the Bowditch
    > > table methods. If you really want to plot this on a UPS what you
    > > describe would be satisfactory.
    > >
    > > Do you have UP sheets for those latitudes?  If not you can construct
    > > your own constant latitude sheet using Lo divisions as cosine of mid lat
    > > in paper dimensions.
    > >
    > > Tony
    > >
    > > Ed Kitchin wrote:
    > > >
    > > > 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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