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    Re: Longhand Sight Reduction
    From: UNK
    Date: 2014 Jun 12, 05:40 +0100

    Thanks Hanno,

    I didn’t understand it all, but sounds like the Bygrave wins cotans down?

    Had a feeling he knew what he was doing!



    From: NavList@fer3.com [mailto:NavList@fer3.com] On Behalf Of Hanno Ix
    Sent: 12 June 2014 05:11
    To: francisupchurch---.com
    Subject: [NavList] Re: Longhand Sight Reduction



    here are the results of a study re: Bygrave error, scale length, etc.

    I was wondering why Bygrave chose his particular sight reduction
    formula and his design. Since nobody on the list answered

    I decided to look into it myself.


    I started out by calculating the lengths of a
    log10(sin(x)) - scale and the length of a log10(tan(x)) - scale

    both with a "1 mm per 1 arcmin"  resolution.

    This resolution was chosen because I assumed it being close to what I need for
    a comfortable reading. However, I will not argue against "2mm per arcmin"

    or thereabouts.
    The lengths came out as:
    log10(sin(x))  64 meters   log10(tan(x)) 28 meters.

    The reason for this large difference is the simple fact that the

    step size of the first varies much more than the one of the second.

    G. LaPook's Flat Bygrave's (FB)  tan() - scale stretch over only ~ 380" or ~ 10 meters.,
    with the cos() - scale about half as far or less. Why, then, can the FB yield - occasionally at least - a result with an error of  2 arc min or less as G. LaPook reports?

    BTW: If the FB were twice as long this error would be, I assume, half of that.

    That chimes roughly with my assumption of "1mm per 1 arcmin" of a readable scale.
    In fact, on the copy of the FB in my possession, at ~45 deg on  the tan() scale you can at discern 1 arcmin with ease - outside much less.

    Getting closer to the real question: In the FB, the cos() scale has been shrunk far below that "1mm per 1 arcmin" threshold. Look at the 5 deg area! Why is the result still as good as Gary et al. claim? In other words, why does that very rough cos() scale not prevent the overall results to be so close to the correct one? That is odd!

    Two components to the answer:

    -  H is read on the scale with the highest resolution, the tan() - scale. ( Actually, that is the case for the azimuth as well.)

    -  In Bygrave's formula, the cos() and its errors have a limited control, so to speak.

    The 2 attached graphs demonstrate this. I will not bother you with the details of the calculations. The first shows the error dH of H given a correct Y and a grossly wrong azimuth A. Its offset dA  is assumed to be as much as 10 arc min. Same for the second graph with dA only 100 arc min. ( Both are interesting cases in their own right.)

    We conclude for


    dA = 10 arcmin: Unless the pair (Y, A)  falls in to red zone dH will never exceed 1 arcmin.

    The good area (white) is a very wide area. Most sight reduction cases will fall into it.

    dA = 100 arcmin: Despite this large error of the azimuth the Bygrave will produce good results for all values of Y if the azimuth itself is below 20 deg. If we however limit the range of admissible Y's  to ~ 20 deg the azimuth may be as large as 50 or 60 deg and the Bygrave will still deliver good results. And for the application of this case: there are often other, perhaps simpler, means to determine the azimuth with an error of  up to 100 arcmin. In those cases, one can take the value found with those and proceed to finding H on the Bygrave.

    I hope this has found your interest.  I admire the insights of Bygrave that he put to use in his great little slide rule!










    On Wed, Jun 11, 2014 at 12:16 PM, Francis Upchurch <NoReply_Upchurch@fer3.com> wrote:

    Thanks Hanno.

    Good challenge. Hospital next week. Out of action maybe 4 weeks, then ok for go. !

    My slide rules cost about $5 a piece so far, but plenty of work!

    Can be done.

    Keep up the interesting work. My money is on Bygrave, but I’m biased!



    From: NavList@fer3.com [mailto:NavList@fer3.com] On Behalf Of Hanno Ix
    Sent: 11 June 2014 17:01
    To: francisupchurch---.com
    Subject: [NavList] Re: Longhand Sight Reduction



    Given the choice Danioli vs Bygrave: What would Chichester have chosen?

    Assuming, of course, he had access to and enough experience with both.


    I celebrated Greg's RIC / Danioli yesterday by playing around with it.
    It indeed works and is fast. Comparison with the standard formula is attached below.

    Greg seems to think one can do that single multiplication with a 10" slide rule.
    I am skeptic. In praxis, ten inchers do not yield correct 4 digits consistently,
    and that's what I need for accuracy over the useful ranges of L,D,t.

    Now, there might be a challenge especially for you: a Fuller < = 10" that can

      - do 4 digit multiplication, i.e. yields vwxy = ABCD * abcd;  v, w, x, y being correct digits.

      - can be built with standard and garage tools plus a PC and printer.

      - in not more than, say, a week.

      - for about $50 or less,  $100 max.


    It need not look like an exhibit in a museum, although it should be sturdy enough
    to survive a 1-week sailing trip in the Virgin Islands.  ( Where and when can I sign up? )

    Re: formulas. One example, with good or flawed results, is not really sufficient to

    judge a formula or method. You need to show it yields accurate 4 digit results consistently
    for the full useful ranges of L,D,t. I am unsure, though, what "useful" means for our CelNav friends.

    Any ideas out there?

    I am studying the Bygrave in this respect right now. Stand by please.



    For the record, Danioli claims:

        sin(h) = n - ( n + m ) * a ;   n: cos(L-D);    m: cos(L+D);   a:  [1 -  cos(t) ] / 2 or hav(t);  

    Let's see. By inserting:

        sin(h) =  cos(L-D)   -  [ cos(L-D) + cos(L+D) ] *  [ 1 - cos(t) ] / 2;

    which is in more detail:

        sin(h) cos(L-D)  - cos(L-D) * [ 1 - cos(t) ] / 2  -  cos(L+D) * [1 - cos(t) ] / 2;

    and more detail yet:

        sin(h) =  cos(L-D)  -  cos(L-D) / 2  +   cos(L-D)*cos(t) / 2   -  cos(L+D) / 2 + cos(L+D)*cos(t) / 2;


         sin(h) =  cos(L-D)/2 - cos(L+D) / 2    +     [ cos(L-D) / 2 + cos(L+D) / 2 ] * cos(t);

                  =         sin(L) * sin(D)              +              cos(L) * cos(D) * cos(t);

    which is correct.


    On Tue, Jun 10, 2014 at 11:11 PM, Francis Upchurch <NoReply_Upchurch@fer3.com> wrote:

    Oh dear. Is it time to put my beloved Bygrave away? Cant wait to here more details of the Bygrave maths.Chichester said he preferred the Bygrave when flying single handed, because he made mistakes with log tables. (Perhaps he did not have Haversines?)  But, could someone explain the main difference/advantages/disadvantages of the versine method (Vers ZD=Vers LHAxCos Latx Cos Dec+Vers(Lat+/-Dec) and the Haversine method? My versine method (Reeds Astro Nav Tables) uses tables of natural and log versines and log cos (total 11 pages).Does not need sines.

    Versine method

    log vers LHA    9.9019

    log cos Lat      9.9177

    log cos Dec     9.9642

    add              29.7838

    Nat Vers of 9.7838=        0.6081

    Lat-Dec=11°13' Nat vers=0.0191. Add= 0.6272=68°6'. =ZD. 90°-68°6'= 21°54'

    Not a lot in it I would say? quicker for me than reduction tables and I understand what we are doing.

    Please correct me and explain the advantages of the Haversine over the versine. (I do not have haversines but do have versines! Where do I get haversines?)

    Bygrave. H=360°-LHA=78°21', co-lat=55°50', y(w)=64°31', X=colat+y(w)=120°21', Y=180°-X=59°39', > Az =76°24'> Hc 21°54'

    No contest! Took a fraction of the time and no mistakes from looking up 4 figure logs etc. And I've got Az (OK done hundreds of Bygrave LOPs and only a couple of Versines!)

    I'll stick to my Bygrave!





    Attached File:

    (img/127997.bygrave da 10.gif: Open and save)

    Attached File:

    (img/127997.bygrave da 100.gif: Open and save)

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