|
|
Subject: Re: Figuring Course given Lat/Long of destination
From: Ed Kitchin (edk@XXX.XXX)
Date: Mon Feb 28 2000 - 18:47:44 EST
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" <severdia@XXX.XXX>
To: <NAVIGATION-L@XXX.XXX>
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" <severdia@XXX.XXX>
> > To: <NAVIGATION-L@XXX.XXX>
> > 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
> > >
>
|