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    Re: The Bygrave Slide Rule
    From: Gary LaPook
    Date: 2009 Jun 02, 04:36 -0700

    This has been very interesting and I want to thank Peter for bringing 
    this up and for getting me thinking more about the Bygrave.
    I decided to investigate his point by proving that the Bygrave has a 
    cosine scale since if the scale is a cosine scale then the other scale 
    must be a cotangent scale. Well what I found is that it is a cosine 
    scale or it is a secant scale because these two scales have identical 
    spacings so the scales are identical. The same thing is true of a 
    tangent and cotangent scale, they are identical. The only difference is 
    that, by convention, they run in opposite directions. You read out a 
    tangent on a conventional slide rule on the "C" scale while, at the same 
    time, you read out the cotangent on the "CI" (C inverted) scale. What 
    this means is that these values are found the same distance from each 
    end of the slide, tangent from the right end and cotangent from the left 
    end. Since the log scale is linear this also shows that the logs of 
    these values are also found the same distance from each end. But the 
    logs of the tangent and cotangent (and also the cosine and secant) if 
    added together always total to one. You can also see this from a log 
    trig table. I have attached the 44º page of table 3 from Bowditch. For 
    example, look at the mantissas for tan and co-tan, .98484 and .01516. 
    Add them together and you get 1.00000. This means that these logs are 
    found exactly the same distance from the ends of the scale, .01516 from 
    the left edge for the co-tan and .01516 from the right edge for the tan. 
    The same goes for the pairs cosine/secant and sine/cosecant.
    Bygrave gives three formulas for the three steps in the process. The 
    first step is tan y = tan Dec / cos LHA and you can solve it this way 
    with a normal slide rule or with a calculator. You can try one for 
    practice, using lat of 57, dec 10 and LHA 25 gives y of 11, Y of 44, Az 
    of 32.47 and Hc of 39º 10'. (Y= 90 -lat + y.)
    But the Bygrave doesn't solve it this way based on the manipulations of 
    the slide rule. On the assumption that it contains a cotangent scale, I 
    wrote previously that the formula is transformed into
    cotan y = co-tan DEC  cos LHA. You can prove this to yourself with an 
    ordinary slide rule, simply reverse the slide containing the tan scale 
    and it becomes a co-tan scale. Do the calculation again and you will get 
    the same answers.
    BUT, the formula can also be transformed into :
    tan y = tan DEC sec LHA and it can be solved with the exact same 
    manipulations as he previous example, the same manipulations as when 
    there is a co-tan scale. Reverse the slide containing the cos scale and 
    it becomes a secant scale.
    The second step has formula:
     tan AZ = cos y tan LHA/cos Y
    This can be rewritten  as cot AZ = cos Y  cot LHA/cos y
    but also as:
    tan AZ = sec Y tan LHA / sec y.
    The Bygrave actually accomplishes this second step by solving either the 
    second or third formula sinc they both require the exact same 
    manipulation. You can also do this on a regular slide rule by again 
    reversing the slide.
    The third step is:
    tan Hc =  cos AZ  tan Y
    but is actually solved with the manipulation of:
    cotan Hc = cot Y/ cos AZ
    or with:
    tan Hc = tan Y / sec AZ
    each requiring the exact same manipulations.
    So, the bottom line is that the Bygrave contains either a  cosine and a 
    co-tan scale or it contanes a tan and a secant scale, take your choice 
    since the two choices are identical requiring identical manipulations.
    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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