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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@POST-GAZETTE.COM]
> Reply To:     Navigation Mailing List
> Sent:         Tuesday, February 29, 2000 10:25 AM
> To:   NAVIGATION-L@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@DREAMSCAPE.COM]
> > Reply To:     Navigation Mailing List
> > Sent:         Monday, February 28, 2000 6:47 PM
> > To:   NAVIGATION-L@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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