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    Re: The Great Circle Challenge
    From: Hanno Ix
    Date: 2014 Dec 28, 17:04 -0800

    Not that it would be important, but you can't really switch d and L in the hav-Doniol
    because doing so makes hav (L +/- d) to hav (d +/- L), and both are equal. So the sum Q
    of them is also not affected by a switch. And these are the only locations
    where d and L appear in the formula. So when you use the hav-DONIOL switching
    d and L does not create meaningful redundancy.

    In another vein: the performance of the BN you observe dazzles me.
    Would it be possible to shoot a close-up of the scale location where it shows Hc?

    The reason it dazzles me is this: I suppose it's range of Hc goes from 0 to 90 deg.
    Therefore, for a resolution of 1 nm you need to discern 5400 different locations on
    some sort of a scale.

    On the Bygrave you have scales exceeding 30 feet of length. On my copy of
    Robin Stuart's flat version the cot-scale - the one which determines the
    resolution of a Bygrave - is 33 feet long, and at the tightest spot at 45 deg you
    have 22 mil ( ~ 0.5 mm)  for 1 nm. This is quite sufficient to make a clear visual distinction.

    I wonder how the BN achieves such distinctions.


    On Sat, Dec 27, 2014 at 11:36 PM, Francis Upchurch <NoReply_Upchurch@fer3.com> wrote:

    Thanks Hanno,

    I’m sure you may be right. In this case however, the errors were different each time and eventually, I got repeats that were consistent, so for different random errors, repeating same process does have some redundancy. I repeated with Doniol, but using the same erroneous LHA and therefore “t or H” and got the same result as per Bygrave.

    Actually, I suspect the multiple errors had more to do with a rather fine 2007 vintage Spanish liquid called Rioja than anything else!

    Morale is , do not drink and drive or navigate!

    PS, with the Bygrave (and probably Doniol?) there is any easy check. If you repeat but interchanging declination and latitude, it gives you same Hc, but wrong Az. So a good redundancy system would be to use your Az graph for Az (or course in a great circle) and then use the Brown-Nassau or Bygrave just for Hc (or (90˚-Hc)x60 for the distance.) In fact, the Brown-Nassau for just Hc +Az from your graph gives the correct result in about  a minute!




    From: NavList@fer3.com [mailto:NavList@fer3.com] On Behalf Of Hanno Ix
    Sent: 28 December 2014 02:06

    To: francisupchurch---.com
    Subject: [NavList] Re: The Great Circle Challenge


    Hi, Francis -

    now, if you hadn't access to this site here - out on the boat all by yourself -
    how would you know you made an error?


    I assume you would established some redundancy in your routines.
    If my life depended on the accuracy of sight reduction I would not sail without one.

    Repeating the same method would not be advisable - not areal redundancy.
    One obvious one would be two people doing two sight reductions

    independently and compare. But even in this case questions arise:

    should they use the same method or different ones? What makes for a better

    error checking?


    BTW: Did the Royal Navy apply redundancy? If so, in what way?
    Numerous examples presented in the classical literature by expert authors

    come up with the wrong answers. You know, complicated sign rules, logarithmic
    calculations, multiple tables etc. plus the appearent absence of redundancy 
    make an excellent recipe for committing errors.

    OK, then: What is a good sequence of redundant sight reduction?

    I think, one may start with the assumption the first one was wrong and then proves
    this assumption of an error being wrong.

    Would you use the Bygrave first and then check with a BN?





    On Sat, Dec 27, 2014 at 4:21 AM, Francis Upchurch <NoReply_Upchurch@fer3.com> wrote:

    Apologies I made an error on the Bygrave. Several check repeats now give 4059 nm distance. 25˚55’ course.



    From: NavList@fer3.com [mailto:NavList@fer3.com] On Behalf Of Francis Upchurch
    Sent: 27 December 2014 09:41
    To: francisupchurch---.com
    Subject: [NavList] Re: The Great Circle Challenge


    Merry Xmas Navlist folk.

    The following probably adversely affected by seasonal excesses.

    Bygrave: Course 25˚55’, dist 4063nm. Took 2 minutes. (substitute Dec= arrival lat, LHA (t) = difference longitude.)

    Brown-Nassau prototype: Course 26˚. Dist 4070nm. Took 1min 30 secs approx.

    What is the best ,correct answer?

    Best wishes,

    Francis Upchurch


    From: NavList@fer3.com [mailto:NavList@fer3.com] On Behalf Of David Pike
    Sent: 26 December 2014 16:21
    To: francisupchurch---.com
    Subject: [NavList] The Great Circle Challenge



    The Great Circle Challenge




    Clearly, from recent posts, we all have a favoured way of solving great circle sailings (which we’ll probably never use for real).  Let’s put them to the test by trying something a bit more complicated like a southern to northern hemisphere crossing combined with a crossing of 180degrees E/W.  How about emulating Captain Cook by travelling from Cook Strait (CS), New Zealand to Waimea Bay (WB), Kauai, Hawaii?  Cook followed the pretty route, but we’ll go direct by great circle.  The coordinates are CS 41d 30’S, 174d30’E to WB 21d57’N, 159d 40’W.  Use your favourite method and report back on your answer, the time it took you, and any difficulties encountered.


    I’ll stick with the diagram method, because at least I’ll know what I’m trying to prove, and I won’t have as many rules to remember and apply which might or might not work.  Dave


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