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    Re: The Bygrave Slide rule
    From: Brad Morris
    Date: 2009 May 29, 13:36 -0400

    I brought out the extended markings to help clarify the discussion!  I 
    realized that tan(89) does not equal the tan (91), yet the value denoted on 
    the scale is the same for either tan(89) and tan(91).  The same applies to 
    the outer scale, there is a sign change at 90 degrees, although the abs 
    values are the same.
    While I remain interested in resolving what the scales actually represent, I 
    don't know that we have enough information yet. Peter has essentially 
    utilized trigonometric identities to "flip the problem over".  Neither 
    argument is compelling enough to state, hey, that's the answer.  No argument 
    accounts for the absolute value, yet that must occur.  The scales are 
    logarithmic, yet there are no logarithms of negative numbers.
    Best Regards
    -----Original Message-----
    From: NavList@fer3.com [mailto:NavList@fer3.com] On Behalf Of NavList@fer3.com
    Sent: Friday, May 29, 2009 5:46 AM
    To: NavList@fer3.com
    Subject: [NavList 8441] The Bygrave Slide rule
    Brad's point is interesting - he points out that the reversed 90-180 degree 
    markings on both scales show that the scales are logs of abs(tan) and 
    abs(cos) respectively. Maybe this is just a 'kludge' for the annoying fact 
    that logs of negative numbers don't exist. It's left to the user to sort out 
    the signs when using logs to work with mixed-sign numbers.
    However, I don't think Brad's point does impact on my postings, since it 
    applies whether or not the scales are 'thought of' as cotan/cos or 
    In effect all I am saying is that I found the references to cotangent scales 
    in the papers by Ron and Gary were counter-intuitive, and the alternative 
    tangent/secant interpretation was easier to grasp, so I thought it worth a 
    posting. If no-one else agrees with me then I will go back into my cave.
    I can't contribute an answer to Gary's puzzle about the apparent impossibility 
    of using the tangent/cosine method at the equinoxes (this puzzle is not 
    limited to the Bygrave solution). A quick look at the celestial triangle 
    tells me that if the sun is near the equator then the perpendicular from the 
    sun to the observers meridian is also near the equator, so the equation for 
    azimuth simplifies to tan(A)=tan(H)/cos(c), but this may already be what Gary 
    has rejected as not being very pretty. I wondered whether the discarded 
    azimuth in the alternative calculation might be a clue, but this is just 
    equivalent to swapping the observer with the sun. The altitude comes out the 
    same of course, but the discarded 'false' azimuth is then seen to be the 
    azimuth of the observer as seen from the sub-solar point, which isn't very 
    useful, even on March 21st.
    Peter M
    [Sent from archive by: peter.martinez-AT-btinternet.com]
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