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    Re: The Bygrave Slide Rule
    From: Gary LaPook
    Date: 2009 May 28, 12:35 -0700
    It works either way, that's the important part.

    I have a question for you that nobody has yet been able to answer for me. For cases where the declination is less than the lowest value on the the tangent/co-tangent scale, Bygrave said to transpose the declination and latitude values and then calculate the altitude which works fine. But he gives no instruction on how to calculate azimuth in this case. I have found a way to closely approximate the azimuth, accurate to less than one degree as I wrote on April 14th:

    So your MHR-1 doesn't provide for cases where the declination is less than 20', the lowest mark on the cotangent scale. Bygrave says to use the same procedure in this case as in the case where the azimuth is near 90?, simply interchange the declination and the latitude and then compute the altitude and this works fine and you get accurate altitudes. But the azimuth that you derive in this process is not the correct azimuth and is thrown away and not used for plotting the LOP just as in the case of azimuths near 90?. Bygrave gives no instruction for computing the azimuth in this case. It cannot be computed in the normal way since the first thing you need to to is find declination on the cotangent scale and values less than 20' are not on the Bygrave or MHR-1. This is an important special case since the sun's declination is in this range for several days around each equinox. I developed an approximation that works well giving azimuths within one degree of he correct value and usually much closer. I simply skip the first step, the derivation of "y" (Bygrave's terminology), "W" (my terminology). I simply set "y" equal to declination and then proceed normally.

    http://groups.google.com/group/NavList/browse_thread/thread/20ac6296103b4c7e?hl=en#

    Although this works fine, I am not happy with the inelegance of this solution. Do you have any information on instructions provided for this case either by Bygrave or on the MHR-1 or a mathematical solution that provides an exact and elegant solution?

    gl

    NavList@fer3.com wrote:
    My thanks to Ron and Gary for responding to my posting.
    
    I should say that I had already read both their papers on this topic. Indeed 
    it was the reference to 'cotangent' in both papers that caused me so much 
    difficulty that I had to dig deeper. I decided to post here after I had sorted out in my own mind how the Bygrave Slide Rule worked. It seemed to me there was a case for questioning the cotangent idea itself.
    
    Both Bygrave's prototype and the production models employ scales on the 
    inner and outer cylinders which have low-numbered degrees at the left end, 
    and high-numbered degrees at the right end of the unwrapped linear 
    equivalent. I think we all understand why it's a good idea to have both degree scales the same way round, and we would probably also all agree that left-to-right increasing is intuitively 'right'.
    
    If we go with Ron and Gary and think of the inner scale as a log cotangent, 
    then the log value of the left end is log(cotan(0)) = +infinity and the log 
    value of the right end is -infinity. That is, decreasing to the 
    right. The same applies to the outer scale if we go with Ron and Gary and 
    consider it as a cosine scale.
    
    I am not suggesting that Ron and Gary are wrong, but that there is an 
    alternative view and that is not wrong either. Indeed it has some good 
    points.
    
    If I think of the inner scale as a log tangent, then the logarithm values it 
    represents increase to the right. If I also think of the outer scale as a log secant, the log values also increase to the right. This is how a conventional slide rule is organised. The numbers and the logarithms both increase left-to-right. This feels intuitively 'nice' to me.
    
    In Ron's paper, he details an experiment in which he aligns the zero degree 
    mark of the outer scale to the 45 degree mark on the inner scale. By noting 
    that the 60 degree mark on the outer scale then aligns exactly with the 63 
    degree mark on the inner scale, and noting that cosine(60) = cotangent(63), he cites this as confirming his assertion that the two scales are indeed log cosine and log cotangent respectively.
    
    However, I would point out that secant(60) = tangent(63), so this same experimental result is consistent with my secant/tangent hypothesis too!
    
    Maybe it boils down to an arbitrary choice of interpretation. We can't even
    go back to Bygrave and ask his help in the argument, since he firmly
    labels the inner scale 'log tangent' and the outer scale 'log cosine' - a
    hybrid of the two alternatives.
    
    My guess is that Bygrave would say that he just wanted to keep the degree
    markings on both scales so that zero was on the left, which is what the user
    would expect. Sticking with tangents and cosines (the only functions named
    in any of the historical documentation), this meant that the log tangents
    increased to the right and the log cosines increased to the left. This
    wasn't a problem: it just meant careful wording of the instructions to 
    produce the required result. He would have seen no need to rename one of the 
    scales, since the scale names don't appear in the instructions anyway.
    
    As I said in my first posting, these instructions, printed clearly on both
    the prototype and production devices, transpose the sequence of operations
    normally used on a conventional slide rule for multiply and divide. They
    appear to multiply the 'cosines' when the equations require division, and
    vice versa. It is this that led me to the realisation that it would be more
    constructive to discard the cotangent/cosine hypothesis and think of the
    cosine scale as a 1/cosine, or secant scale.
    
    Peter Martinez
    
    --------------------------------------------------------
    [Sent from archive by: peter.martinez-AT-btinternet.com]
    
    
    
    
      


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