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    Re: Figuring Course given Lat/Long of destination
    From: Joe Shields
    Date: 2000 Feb 29, 8:26 AM

    Forget Hs and Hc and any sight corrections.  Forget your sextant.  Forget
    what time it is. This is just dealing with the theoretical side of things
    starting with latitude, declination, and  local hour angle (LHA).  Instead
    of looking up GHA and Declination from your Nautical Almanac (or whatever)
    and then computing LHA from the diff between GHA and your starting
    Longitude, just take the difference between starting long. and destination
    long. to get LHA.  Substitute destination latitude for Declination, and then
    do the trig. or sight reduction.  Of course part of the problem might be HO
    249 which makes you use an AP (assumed position) instead of allowing a DR
    lat/long.  I use HO 211, which allows me to use my actual lat/long.  At any
    rate, the point still needs to be emphasised, that this will only give you a
    bearing to start off with.  To continue following a great circle route, you
    would need to recompute a Zn from other intermediate points along the way,
    creating waypoints.
    
    -- Joe Shields
    
    > ----------
    > From-         Ed Kitchin[SMTP:edk{at}DREAMSCAPE.COM]
    > Reply To:     Navigation Mailing List
    > Sent:         Monday, February 28, 2000 6:47 PM
    > To:   NAVIGATION-L{at}LISTSERV.WEBKAHUNA.COM
    > Subject:      Re: Figuring Course given Lat/Long of destination
    >
    > 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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