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    Re: Great Circle Initial Course Boston to Wellington?
    From: Stan K
    Date: 2017 Dec 5, 12:37 -0500

    In going over my old Power Squadrons course materials, an appendix shows two methods of determining GC waypoints:  A) According to Increments of Longitude; B) According to Increments in Miles.  (A being the way I did it, B being the way you did it.)  But the course itself only taught Method A, and the reason was the way it was taught:  "The JN Course...requires Great Circle courses to be drawn on a gnomonic projection and then transferred to a Mercator chart.  Latitudes are determined from the gnomonic chart at points crossing longitudes divisible by 5.  These were chosen because 5º longitude increments are printed as solid lines."

    So both methods were apparently in common use.  I'd be tempted to add the second method to Celestial Tools.  However, note that I use past tense in the above paragraph, as neither way is taught in the current Power Squadrons courses.  Still, the temptation is there.  At this time I think I'll just put a note about Method B in the Help.

    For the proposed reasons why the waypoints stopped prematurely, definitely go with the former! ;-)


    -----Original Message-----
    From: Sean C <NoReply_SeanC@fer3.com>
    To: slk1000 <slk1000@aol.com>
    Sent: Tue, Dec 5, 2017 9:22 am
    Subject: [NavList] Re: Great Circle Initial Course Boston to Wellington?

    My spreadsheet uses the law of cosines formulae (the same ones used for sight reduction) to find the total distance, initial course, waypoints and intermediate courses. I can't remember where I originally read about it, but the tabular method is explained in Section D: Other Aplications of Pub. 229 and also in the 2002 ed. of Bowditch, Chapter 24: The Sailings. This can, of course, be adapted for use with the formulae quite easily. Following are instructions on how to do just that...
    To find G.C. distance and initial course:
    Substitute the declination with the latitude of the destination and the L.H.A. with the difference in longitude. Thus, we have:
    asin(sin(φ₂) ∙ sin(φ₁) + cos(φ₂) ∙ cos(δλ) ∙ cos(φ₁)) = D ; 60 ∙ (90 - D) = G.C. distance
    acos((sin(φ₂) ∙ cos(φ₁) - cos(φ₂) ∙ cos(δλ) ∙ sin(φ₁)) / cos(D)) = C ; If δλ < 180°, course = 360° - C; otherwise course = C
    To find waypoints along the G.C. route:
    To find the latitude and difference in longitude of the waypoint: substitute the declination with 90° - distance to the waypoint (e.g. use 5° for legs of 300 NM) and substitute the L.H.A. with the initial course. thus:
    asin(sin(85°) ∙ sin(φ₁) + cos(85°) ∙ cos(I.C.A.) ∙ cos(φ₁)) = latitude of the waypoint 300 NM from the departure point
    acos((sin(85°) ∙ cos(φ₁) - cos(85°) ∙ cos(I.C.A.) ∙ sin(φ₁)) / cos(φ wpt.)) = difference in longitude of the waypoint from the departure point
    To find the next waypoint (600 NM from the departure point), one would simply substitute 85° with 80°. The I.C.A. remains unchanged for all of the waypoints. As you realized, this method yields waypoints which are 300 NM apart along the G.C.course, but not necessarily 5° of longitude apart. This has the advantage of keeping each leg of the route the same distance no matter what angle at which the course cuts the meridians. The "distance remaining" column in my spreadsheet was a quick and dirty solution - subtracting 300 NM from the total distance for each waypoint. I didn't really check to see if that was absolutely accurate. To find the intermediate courses, one simply needs to run through the first set of equations again with the waypoints as the departure and destination.
    As for why the waypoints stop with more than 300 NM left ... well, that was either due to a limitation of Excel or my ineptitude at getting it to do what I wanted or a little of both. I can't quite remember. I'm going with the former, though, because that sounds better to me. :D
    Seriously, though, I'll have to go back and look at it and see if I can't figure out how to fix that.
    Sean C.
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