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    Oops, forget what I said about forgetting Hc
    From: Joe Shields
    Date: 2000 Feb 29, 8:40 AM

    oops, forget what I said about forgetting Hc.  With the resulting
    (calculated) Hc, you can subtract it from 90 to get the distance in nautical
    miles from source to destination (each degree = 60 nm, each minute = 1 nm).
    
    -- Joe
    
    
    > ----------
    > From-         Joe Shields[SMTP:jshields{at}POST-GAZETTE.COM]
    > Reply To:     Navigation Mailing List
    > Sent:         Tuesday, February 29, 2000 10:25 AM
    > To:   NAVIGATION-L{at}LISTSERV.WEBKAHUNA.COM
    > Subject:      Re: Figuring Course given Lat/Long of destination
    >
    > 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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